Question

question 12 I need a full solution

Answer 1

Question : If the sum of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is 0.

Proof :

Let, a be the first term of the A.P. and d be the common difference.

• Sum of m terms of the A.P.

= m/2 * [2a + (m - 1)d]

• Sum of n terms of the A.P.

= n/2 * [2a + (n - 1)d]

By the given condition,

sum of m terms = sum of n terms

⇒ m/2 * [2a + (m - 1)d] = n/2 * [2a + (n - 1)d]

⇒ m {2a + (m -1)d} = n {2a + (n - 1)d}

⇒ 2am + (m - 1)md = 2an + (n - 1)nd

⇒ 2a (m - n) + {(m - 1)md - (n - 1)nd} = 0

⇒ 2a (m - n) + (m² - m - n² + n)d = 0

⇒ 2a (m - n) + {(m² - n²) - (m - n)}d = 0

⇒ 2a (m - n) + {(m + n)(m - n) - (m - n)}d = 0

⇒ 2a (m - n) + (m - n)(m + n -1)d = 0

⇒ 2a + (m + n - 1)d = 0 .....(i)

∴ the sum of (m + n) terms of the A.P.

= (m + n)/2 * [2a + (m + n - 1)d]

= (m + n)/2 * 0 , by (i)

= 0

Thus, proved.