Question

sum of 3rd and 7th term of a A.P is 6 and their product is 8 find the sum of first 16th terms of A.P

Answer 1

☞ Sum of 3rd and 7th term of a A.P is 6.

$a_{n}$ = a + (n - 1)d

$a_{3}$ = a + (3 - 1)d

$a_{7}$ = a + (7 - 1)d

A.T.Q.

=> a + (3 - 1)d + a + (7 - 1)d = 6

=> a + 2d + a + 6d = 6

=> 2a + 8d = 6

=> a + 4d = 3

=> a = 3 - 4d _________(eq 1)

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☞ And their product is 8.

The product of above numbers i.e.

of.. (a + 2d) and of (a + 6d) is 8.

=> (a + 2d) (a + 6d) = 8

=> (3 - 4d + 2d) (3 - 4d + 6d) = 8

[From (eq 1)]

=> (3 - 2d) (3 + 2d) = 8

=> 3 (3 + 2d) [(-2d) (3 + 2d)] = 8

=> 9 + 6d - 6d - 4d² = 8

=> 9 - 4d² = 8

=> - 4d² = - 1

=> 4d² = 1

=> d² = 1/4

=> d = 1/2

• Put value of d in (eq 1)

=> a = 3 - 4(1/2)

=> a = 3 - 2

=> a = 1

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☞ We have to find the sum of first 16th terms of an A.P.

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

Here,

n = 16

a = 1

d = 1/2

$S_{16}$ = $\dfrac{16}{2}$ [2a + (16 - 1)d]

=> 8 [2(1) + 15$(\dfrac{1}{2} )$

=> 8 (2 + $\dfrac{15}{2})$

=> 8 $(\dfrac{19}{2})$

=> 76

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Sum of first 16th term is 76.

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