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Given: S₃ + S₈ = 7

S₇ + S₁₄ = -3

To find: a₁₀

Formula Used:

Sum of “n” terms of an AP:

Where S_n is the sum of first n terms

nth term of an AP is:

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

Where a_n is the last term.

n = no of terms

a = first term

d = common difference

Explanation:

Let the first term and common difference of an AP are a and d, respectively.

a₃ + a₈ = 7

As we know, nth term of an AP is

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

where a = first term

a_n is nth term

d is the common difference

a + 2d + a + 7d = 7

2a + 9d = 7

2a = 7 - 9d [ Eqn 1]

a₇ + a₁₄ = - 3

a + 6d + a + 13d = - 3

2a + 19d = - 3

7 - 9d + 19d = - 3 [ using eqn i]

7 + 10d = - 3

10d = - 10

d = - 1

using this value in eqn i

2a = 7 - 9 (- 1 )

2a = 16

a = 8

Now,

a₁₀ = a + 9d

= 8 + 9 (- 1 )

= 8 - 9 = - 1