Question

The sum of first 31 terms of an arithmetic sequence is 620,
(a) What is its 16th term ?
(b) What is the sum of 15th and 17th terms?
(c) Find the sum of first and 31st terms.​

Answer 1

Hope this explaination helps

Answer 2

Solution :-

Let ,

• first term of given AP = a
• common difference = d .

so,

→ Sn = (n/2)[2a + (n - 1)d]

→ 620 = (31/2)[2a + (31 - 1)d]

→ (620*2)/31 = 2a + 30d

→ 40 = 2(a + 15d)

→ a + 15d = 20 ---------- Eqn.(1)

then,

→ T(n) = a + (n - 1)d

→ T(16) = a + (16 - 1)d

→ T(16) = a + 15d

putting value from Eqn.(1),

→ T(16) = 20 (Ans.a)

also,

→ T(15) + T(17) = {a + (15 - 1)d} + {a + (17 - 1)d}

→ T(15) + T(17) = (a + 14d) + (a + 16d)

→ T(15) + T(17) = (2a + 30d)

→ T(15) + T(17) = 2(a + 15d)

putting value from Eqn.(1),

→ T(15) + T(17) = 2 * 20

→ T(15) + T(17) = 40 (Ans.b)

and,

→ T(1) + T(31) = a + {a + (31 - 1)d}

→ T(1) + T(31) = a + (a + 30d)

→ T(1) + T(31) = (2a + 30d)

→ T(1) + T(31) = 2(a + 15d)

again, putting value from Eqn.(1),

→ T(1) + T(31) = 2 * 20

→ T(1) + T(31) = 40 (Ans.c)

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