#### Progressions

1)

What is the sum of the given series?
13, 17, 21…77

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The given series is in A.P. First term a = 13, Common difference = 17 – 13 = 4 tn = a + (n-1)d à 77 = 13 + (n-1)4 => n=17 Sn = $${n \over {2}}$$(a+l) S17 = $${17 \over {2}}$$(13+77) = 765.Hence (a).

2)

The 25th term and the 30th term of an AP are 90 and 140 respectively. What is the 35th term?

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By property
Equidistant terms of an AP are also in AP
( t n – t n -k )   = ( t n+k – t n )
(t 30 –t 25 )  = ( t 35 –t 30 )
140 – 90 =  t 35  – 140
t 35 = 190  Hence (b).

3)

Find the 12th term in the given series 17, 22, 27...

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The given series is in A.P.
t n = a + (n- 1)d
t 12 = 17 + (12-1)5 = 72 Hence (d).

4)

Find the sum of the first 5 terms of the G.P 3, – 12, 48, – 192….

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a=3, r =  $${-12 \over {3}}$$ = -4
s n = $${a(1-r^{n}) \over {1-r}}$$
s 5 = $${3(1-(-4)^{5}) \over {1-(-4)}}$$ = 3 * 205 = 615 .  Hence (c).

5)

What is the 72th term of the G.P whose 24th and 48th terms are 180 and 540 respectively?

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Common ratio r =  $${t{}{}{}{}{}{}{}{}{}{}{}{} \over {t}}$$ =  $${t \over {t}}$$ =  $${t \over {t}}$$ =  $${t \over {t}}$$
=>$${540 \over {180}}$$ =  $${t \over {540}}$$
=> 1620 Hence (b).

6)

The common ratio of a G.P is $${-2 \over {3}}$$ and the sum to infinity is $${27 \over {5}}$$. Find the first term.

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Sum to infinity of a GP = $${a \over {1-r}}$$
$${a \over {1-({-2 \over {3}})}}$$ =  $${27 \over {5}}$$
a = 9  Hence (c).

7)

Find the number of terms in the series
$${1 \over {9}}$$, $${1 \over {16}}$$, $${1 \over {23}}$$,…$${1 \over {114}}$$

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a=9, d = 7 tn = $${1 \over {a + (n-1)d}}$$ $${1 \over {114}}$$ = $${1 \over {9+(n-1)7}}$$à n=16.Hence (b).

8)

Find the 11th term of the given series
$${1 \over {7}}$$, $${1 \over {11}}$$, $${1 \over {15}}$$,…

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a =7, d = 4
t 11 =  $${1 \over {7 + (11-1)4}}$$ = $${1 \over {47}}$$ .  Hence (c).

9)

The sum of three numbers in A.P. is 27 and their product is 648. What is the square of third term?

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Let three numbers be (a - d), a, (a + d). (a – d) + a + (a + d) = 27 à a = 9 (a - d) * a * (a + d) = 648 (92d2) * 9 = 648, d = 3. Three numbers are 6, 9, and 12. Square of third number = 122 = 144.Hence (d).

10)

Insert 3 terms in an A.P between 3 and 47.

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Common difference d =  $${b-a \over {k+1}}$$
=>$${47 - 3 \over {3 + 1}}$$ = 11
Therefore, the series is 3,  14, 25, 36 and  47. Hence (a).

11)

What is the sum of all the 3 digit numbers that leave a remainder of ‘6’ when divided by 7?

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The smallest 3 digit number that leaves a remainder of 6 when divided by 7 is 111 Next number is 118. The Largest 3 digit number that leaves a remainder of 6 when divided by 7 is 993 tn = a + (n-1)d 993 = 111 + (n-1)7 à n = 127 S127 = $${127 \over {2}}$$(111+993) = 70104.Hence (d).

12)

In an A.P, t7 : t11 is 11: 13 then find t8 : t12.

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Given t : t 11  = 11: 13
$${a+ 6d \over {a + 10d}}$$ =  $${11 \over {13}}$$
a = 16d
$${t \over {t}}$$ =  $${a+ 7d \over {a + 11d}}$$ =  $${23 \over {27}}$$ Hence (c).

13)

The sum and the product of three terms in G.P are 49 and 2744 respectively. Which of the following can be the first term?

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Let the 3 terms be a/r, a, ar .
Sum of the 3 terms = (a/r) + a + ( ar ) =49 … (1)
Product of the 3 terms = (a/r) * a * ( ar ) = 2744 … (2)
By solving the equation (2), a = 14
Substituting the value of  a in the first equation, we get  r = 2 (or) 1/2
Ther efore, the first term can be 7 (or)  14 . Hence (a).

14)

A man saves money with an initial amount of Rs.3 per day. Next day he saves double the previous day and so on. If he saves money for a total of 11 days, what would be the total amount saved by him?

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Saving s  doubles . Then,  savings in successive days = 3, 6,  12…
This is a GP series. a=3, r=2
S 11 = $${3(2^{11}-1) \over {2-1}}$$ = 6141 .  Hence (b).

15)

If x, 3x + 4, 7x + 12… are in G.P, then which of the following can be the series

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Common ratio r =  $${3x+4 \over {x}}$$ =  $${7x+12 \over {3x+4}}$$
($$3x+4)=x(7x+12)$$
X = -4 or -2.
Therefore the series is  -4, -8, - 16,…   Hence (d).

16)

Insert 5 terms in an G.P between 4 and 2916.

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r =$$({b \over {a}}$$) 1/(k+1) r =$$({2916 \over {4}}$$) 1/(5+1)à r = 3 Therefore, the series is 4, 12, 36, 108, 324, 972 and 2916. Hence (a).

17)

In H.P, the 7thand 9th terms are $${1 \over {36}}$$ and $${1 \over {45}}$$ respectively. Find the 17thterm.

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7 th term of HP =  $${1 \over {a + 6d}}$$ = $${1 \over {36}}$$ , a + 6d = 36
9 th term of HP =  $${1 \over {a + 8d}}$$ =  $${1 \over {45}}$$ , a + 8d = 45
On solving these two equations, a = 9, d = $${9 \over {2}}$$
17 th term =  $${1 \over {a + 16d}}$$ = $${1 \over {9+ 16({9 \over {2}})}}={1 \over {81}}$$ .  Hence (c).

18)

Find the sum all numbers between 1000 and 5000 which are divisible by 7 and 3.

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The numbers between 1000 and 5000 which are divisible by 7 and 3 are 1008, 1029,…4998 tn = a + (n-1)d 4998 = 1008 + (n-1)21à n = 191 S191 = $${191 \over {2}}$$(1008+4998) = 573573. Hence (b).

19)

Find the value of [(p - q + 2r) * (p– q + r)], if p, q, r are in A.P.

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Since p, q, r  are in AP,
2q = p +r => p = 2q – r
[(p - q +2r ) *  (p – q + r )]  ------(1)
Substitute p value in equation (1)
[(p - q + 2r ) *  (p – q + r)] = [(2q - r - q +2r ) *  (2q – r – q + r)]
= q (q + r)
= q 2 + qr.  Hence (d).

20)

What is the sum of all the 3 digit numbers that leave a remainder of ‘3’ when divided by 5 and 6.

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The smallest 3 digit number that leaves a remainder of 3 when divided by 5 and 6 is 123 Next number is 153. The Largest 3 digit number that leaves a remainder of 3 when divided by 5 and 6 is 993 tn = a + (n-1)d 993 = 123 + (n-1)30 à n = 30 S30 = $${30 \over {2}}$$(123+993) =16740.Hence (a).

21)

How many terms are identical in the two APs 7, 14, 21, 28,t32 and 4, 8, 12, …t58?

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The first A.P is a series of the  multiples of 7 and the second  A.P is a series of the multiples of 4 .
The identical terms of the two given series are even
multiples of 28 .
So, the series formed is 28,  56,… 224
Here, a = 28 and d = 28
t n = a + (n- 1)d
224 = 28 + (n – 1)28
n =  8 . Hence (d).

22)

The sum of the first 24 terms of a GP is equal to the sum of the first 26 terms in the same GP. If the sum of the first 33 terms is 98, what is the first term in the series?

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Sum of first 24 terms of the GP = Sum of the first 26 terms  of the GP
So term 25 + term 26 = 0
Take 19 th term = a, 20 th term =  ar
a +  ar = 0
a (1 + r) = 0
Since a ≠ 0, r = – 1
So, the series is a, – a, a, –a ….
Now, S 33 = 98 (given in question )
s n = $${a(r^{n}-1) \over {r-1}}$$
s 33 = $${a((-1)^{33}-1) \over {-1-1}}$$ = 98
∴ The first term a = 98 .  Hence (c).

23)

Find the sum of the series 5 + 55 + 555 + … + 5555 …. 555 up to 35 terms..

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s 35  = 5(1 + 11 + 111+….)
s 35 =  $${5 \over {9}}$$ ( 9 +99 + 999+….)
= $${5 \over {9}}$$ ((10 - 1 ) +(100 – 1) + (100 –  1)+ ….)
= $${5 \over {9}}$$ (( 10 +100 + 1000+….)-35)
= $${5 \over {9}}$$ ( $${10(10-1) \over {9}}$$ - 35)
= $${5(10-325) \over {81}}$$ Hence (a).

24)

Given X = $${4 \over {3}}$$ and Y = (474+473+…+40), which of the following is true?

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Y is in G.P. with a = 4 0 , r = 4, n = 75
s n = $${a(r^{n}-1) \over {r-1}}$$
s 75 = $${1(4^{75}-1) \over {4-1}}$$
= $${4^{75} \over {3}}$$ -  $${1 \over {3}}$$
Y = X - $${1 \over {3}}$$ Hence (c).

25)

If 9, 6, p form a G.P and 2, p, q, r… form a H.P. then, find the 20th term of this H.P. series

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9, 6, p are in GP Where, a = 9 , r = $${2 \over {3}}$$ , p = 6 * ( $${2 \over {3}}$$ ), p = 4 9, 6, 4 are in GP 2, p, q, r…. are in HP 2, 4,…. w here, $${1 \over {a}}$$ = 2 , $${1 \over {a+d}}$$ = 4 t 20 = $${1 \over {2 + (20-1)2}}$$ = t n = 40. Hence (a).

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