Question 1

Find the 10th term of the AP 13, 8, 3, -2,.......?

Accepted Answer

In the given AP, we have a = 13 , d = (8 - 13) = -5 and n = 10 Tn = a + (n - 1)d T10 = 13 + (10 - 1) - 5 = 13 + 9 x (-5) = 13 - 45 = -32

Question 2

What term of the AP 10, 8, 6, 4, .... is - 28?

Accepted Answer

We have, a = 10 , d = (8 - 10) = -2, Tn = - 28 Tn = a + (n - 1)d ⇒ -28 = 10 + (n - 1) x (- 2) ⇒ -28 = 10 - 2n + 2 ⇒ 2n = 12 + 28 ⇒ 2n = 40 ∴ n = 20

Question 3

The 9th term exceeds the fifth term of an AP by 32. The sum of the ninth and fifth terms is 114. Find the eight term of the AP.?

Accepted Answer

T9 = a + (9 - 1)d = a + 8d T5 = a + (5 - 1)d = a + 4d T9 - T5 = 32 ⇒ (a + 8d) - (a + 4d) = 32 ⇒ 4d = 32 ∴ d = 8 T9 + T5 = (a + 8d) + (a + 4d) = 2a + 12d T9 + T5 = 114 ⇒ 114 = 2(a + 6d) ⇒ a + 6d = 57 ∴ T8 = a + 7d = a + 6d + d = 57 + 8 = 65

Question 4

What is the 11th term of the series 1, 8, 15, ...?

Accepted Answer

This is an AP with a = 1 and d = 7 ∴ 11th term = a + (n - 1) x d = 1 + (11 - 1) x 7 = 1 + 10 x 7 = 71

Question 5

Find the value of 1 - 2 - 3 + 2 - 3 - 4 + 3 - 4 - 5 + ... + 100 terms .?

Accepted Answer

The first 100 terms of this series is (1 - 2 - 3) + (2 - 3 - 4) + ... + (33 - 34 - 35) + 34 The first 33 terms of the above series will given an AP as -4, -5, -6, ... -36 ∴ Answer = 33 x (-20 ) + 34 = - 626

Question 6

Find the sum of all numbers, which are divisible by 8 between 200 to 400.?

Accepted Answer

The least number between 200 and 400, which is divisible by 8 is 200. The last term less than 400, which is divisible by 8 is 400 . ∴ The series is 200, 208, 216,... , 400. The number of terms in the AP is (400 - 200/ 8) + 1 = 26 n = 26 d = 8 ∴ Sum = n/2[2a + (n - 1) d] = 26/2 [2x 200 + (26 -1) x 8] = 13[400 + 200] = 7800

Question 7

How many terms of the series -20, -16, -12, ... must be taken so that the sum is 120?

Accepted Answer

Sn = 120, a = -20, d = 4 Sn = n/2[2a + (n - 1)d ] ⇒ 120 = n/2 [2 x (-20) + (n - 1) x 4] ⇒ 120 = -20n + 2(n -1)n ⇒ 120 = -20n + 2n2 - 2n ⇒ 120 = 2n2 - 22n ⇒ 2n2 - 22n - 120 = 0 ⇒ n2-11n - 60 = 0 ⇒ n2 - 15n + 4n - 60 = 0 ⇒ n(n - 15) + 4(n - 15) = 0 ⇒ (n + 4) (n - 15) = 0 ⇒ n = -4 or 15 ∴ n = 15

Question 8

Find the sum of even number below 1672 (including it).?

Accepted Answer

Sum of even numbers = n/2[(n/2) + 1)] = (1672/2) x [(1672/2) + 1] = 699732

Question 9

Find the sum of even numbers below 281.?

Accepted Answer

As, n is odd. ∴ Sum of even number = [(n - 1)/2 ] [ (n + 1 )/2] = 280/2 x 282/2 = 19740

Question 10

The sum to 12 terms of an AP is equal to the sum to 18 terms. What will be the sum to 30 terms for this series?

Accepted Answer

As, S12 = S18 So, S11 = S19 S10 = S20 S9 = S19 ......... ......... ......... S0 = S30 As, S0 = 0 So, S30 = 0

Question 11

The number of bacteria in a certain culture doubles every hour. If there were 50 bacteria present in the culture originally, how many bacteria will be born in 12th hour?

Accepted Answer

The number of bacteria at any given time forms a GP whose terms are given by 50, 100, 200, ... where a = 50, r = 100/50 = 2 Tn = arn -1 ⇒ T12 = 50(2)12 - 1 = 50 x (2)11 = 102400 So, the number of bacteria born in 12th hour is 102400.

Question 12

For doing a certain job, boy is offered ₹ 5 on the first day, ₹ 15 on the second day, ₹ 45 on the third day and so no. How much will the boy earn at the end of seven days?

Accepted Answer

This is a GP with a = 5 , r = 15/5 = 3, n = 7 Sn = a(rn - 1 )/(r - 1) ⇒ S15 = 5(37 - 1)/(3 - 1) = 5/2(37 - 1) = 5/2 (2187 - 1) = ₹ 5465

Question 13

The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4√ 3 and their harmonic mean is 6, then what is the father's age?

Accepted Answer

Given, F = 2E + 4Y ...(I) √EY = 4√3 ⇒ EY = 48 ...(II) and 2EY/ E + Y = 6 ⇒ E + Y = 16 ...(III) Now, (E -Y)2 = (E + Y)2 - 4EY = (16)2 - 4 x 48 = 256 - 192 = 64 ∴ E - Y = 8 ...(iv) From Eqs. (iii) and (iv), we get E = 12 and Y = 4 From Eq. (i) F = 2 x 12 + 4 x 4 = 40 yr

Question 14

The fifth term of a GP is 81 and first term is 16. Find the 4th term.?

Accepted Answer

The nth term of a GP is arn - 1 , 5th term = ar5 - 1 = ar4 = 81 1st term = a = 16 ∴ r4 = 81/16 ⇒ r = ∜81/16 = 3/2 ∴ 4th term = ar4 - 1 = ar3 = 16 x 3/2 x 3/2 x 3/2 = 54

Question 15

How many terms are there in the GP 5, 20, 80, 320, ..., 20480?

Accepted Answer

Let term = l = arn - 1 a = 5 and l = 20480 r = 20/5 = 4 ∴ 20480 = 5 x (4)n-1 (4)n-1 = 20480/5 = 4096 = (4)6 n - 1 = 6 ∴ n = 7

Question 16

A and B are two number whose AM is 25 and GM is 7. Which of the following may be a value of A?

Accepted Answer

(a + b)/2 = 25 a + b = 50 √ab = 7 ab = 49 Hence, A can either be 7 or 49. So, 49 is the answer.

Question 17

If the second term of an HP is 1/6 and the fourth terms is 1/14, then find the tenth term of the HP?

Accepted Answer

The second and fourth term of the HP are 1/6 and 1/14 respectively, Hence, for the corresponding AP, the second term is 6 and fourth term is 14. Hence a + d = 6 and a + 3d = 14 ∴ 2d = 8 ⇒ d = 4 and a = 2 Hence, the 10th term of this AP = a + (10 -1)d = 2 + (10 - 1)x 4 = 2 + 9 x 4 = 2 + 36 = 38 Hence, for the corresponding HP, the 10th term is 1/38.

Question 18

If five times the fifth term of an AP is equal to seven times the seventh term of the AP, then what is the twelfth term?

Accepted Answer

T5 = a + 4d, T7 = a + 6d ∴ 5(a + 4d) = 7(a + 6d) ⇒ 5a + 20d = 7a + 42d ⇒ a = -11d ⇒ T12 = a + 11d = - 11d + 11d = 0 So, the twelfth term is 0.

Question 19

2, 8, 18,?, 50, 72

Accepted Answer

Series pattern 2 x 12, 2 x 22, 2 x 32, 2 x 42, 2 x 52, 2 x 62 ∴ Missing term = 2 x 42 = 32

Question 20

2, 9, 28, 65, 126, .....?

Accepted Answer

Series pattern 13 + 1, 33 + 1, 43 + 1, 53 + 1 ∴ Missing term = 63 + 1 = 217

Odd Man Out and Series Questions