Question

The sum of 8th term and 23rd term of an arithmetic sequence is 75.
a). first term+30th term =_______________
b). first term +26th term =______________
c) find the sum of the first 30 terms.​

Answer 1

Solution :-

we know that,

• Tn(nth term of an AP) = a + (n - 1)d where a is first term and d is common difference .
• Sn(sum of n terms) = (n/2)[first term + last term]

so, let first term of given AP is a and common difference is d .

given that,

→ T(8) + T(23) = 75

→ [a + (8 - 1)d] + [a + (23 - 1)d] = 75

→ [a + 7d] + [a + 22d] = 75

→ (2a + 29d) = 75 ---------- Eqn.(1)

now,

(a) first term + 30th term

→ a + [a + (30 - 1)d]

→ a + (a + 29d)

→ (2a + 29d)

putting value from Eqn.(1) we get,

→ 75 (Ans.)

(b) Fifth term + 26th term

→ [a + (5 - 1)d] + [a + (26 - 1)d]

→ (a + 4d) + (a + 25d)

→ (2a + 29d)

putting value from Eqn.(1) we get,

→ 75 (Ans.)

(c) Sum of the first 30 terms :-

→ T(30) = a + (30 - 1)d = a + 29d

→ S(n) = (n/2)[first term + last term]

→ S(30) = (30/2)[a + a + 29d]

→ S(30) = 15(2a + 29d)

putting value from Eqn.(1) we get,

→ S(30) = 15 * 75

→ S(30) = 1125 (Ans.)

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