Question

If 5th term of an a.p is double the 7th term, prove that the sum of the first 17 term is zero.

Answer 1

Answer:

Step-by-step explanation:

Given,

5th term of an AP = double the 7th term

To prove

Sum of the first 17term = 0

Recall the concept

nth term of an AP = aₙ= a+(n-1)d

Sum to n terms of an AP = Sₙ = $\frac{n}{2}[2a+(n-1)d]$, where a is the first term, and d is the common difference of the AP

Solution:

Since 5th term of an AP = double the 7th term, we have

a₅ = 2a₇

a+4d = 2(a+6d)

a+4d = 2a + 12d

2a -a = 4d - 12d

a = -8d ---------------(1)

Sum of the first 17 terms of the AP = S₁₇ = $\frac{17}{2}[2a+16d]$

Substituting the value of 'a' from equation (1) we get,

S₁₇ = $\frac{17}{2}[2(-8d)+16d]$

= $\frac{17}{2}[-16d+16d]$

= $\frac{17}{2}X0$

= 0

Sum of first 17 terms of the AP = 0

Hence proved

#SPJ2