Q) The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

Ans: Let’s consider the first terms of AP is a and the common difference is d.

(i). Calculating first term & Common difference:

Since nth terms of an AP is given by, T_n = a + (n - 1) d

∴  8th term, T_8 = a + (8 – 1) d = a + 7 d

Since it is given that the sum of first term & 8th terms is 32,

∴  a + (a + 7 d) = 32

∴  a + 7 d = 32 – a….. (i)

Also, it is given that the product of first term & 8th terms is 60,

∴  a (a + 7 d) = 60 …..  (ii)

By solving equation (i) and equation (ii), we get:

a [32 – a)] = 60

∴ 32 a  – a 2 = 60

∴  a 2 – 32 a +  60 = 0

∴  a 2 – 30 a – 2 a +  60 = 0

∴  (a – 30) (a – 2) = 0

Therefore, a = 30 or a = 2

By substituting value of a = 30 in equation (i), we get:

a + 7 d = 32 – a

∴ 7 d  = 32 – 2 a

∴  7 d  = 32 – 2 (30) = 32 – 60 = – 28

∴  d  = – 4

By substituting value of a = 2 in equation (i), we get:

a + 7 d = 32 – a

∴ 7 d  = 32 – 2 a

∴  7 d  = 32 – 2 (2) = 32 – 4 = 28

∴  d  = 4

Therefore, for  first term a = 30, the common difference, d is – 4 and for first term, a = 2, the common difference, d is 4.

(ii). Calculating sum of 20 terms:

We know that the sum of n terms of an AP is given by, S_n = \frac {n}{2} [ 2 a + (n - 1) d]

a) Here, n = 20, a = 30, d = – 4

S_{20} = \frac {20}{2} [ 2 (30) + (20 - 1) (- 4)]

S_{20} = 10 (60 – 76) = – 160

b) Here, n = 20, a = 2, d = 4

S_{20} = \frac {20}{2} [ 2 (2) + (20 - 1) ( 4)]

S_{20} = 10 (4 + 76) = 800

Therefore, for first term a = 30, the sum of 20 terms is – 160 and for first term, a = 2, the sum of 20 terms is 800.

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