Question

If 12^th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Answer 1

Let us take the first term as a and the common difference as d.

Given,

a₁₂ = -13 S₄ = 24

Also, we know that a_n = a + (n – 1)d

So, for the 12th term

a₁₂ = a + (12 – 1)d = -13

⟹ a + 11d = -13

a = -13 – 11d  …. (1)

And, we that for sum of terms

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Here, n = 4

S₄  = \(\frac{4}{2}\)[2(a) + (4 − 1)d]

⟹ 24 = (2)[2a + (3)(d)]

⟹ 24 = 4a + 6d

⟹ 4a = 24 – 6d

Subtracting (1) from (2), we have

Further simplifying for d, we get,

On substituting the value of d in (1), we find a

a = -13 – 11(-2)

a = -13 + 22

a = 9

Next, the sum of 10 term is given by

S₁₀  = \(\frac{10}{2}\)[2(9) + (10 − 1)(−2)]

= (5)[19 + (9)(-2)]

= (5)(18 – 18) = 0

Thus, the sum of first 10 terms for the given A.P. is S₁₀ = 0.