Question

.If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m n) terms is zero

Answer 1

If the sum of first m terms of an AP is the same as the sum of first n terms then the sum of its first (m+n) is 0.

• Sum of AP formula is given by $S_n$ = $\frac{n}{2}(2a + (n-1)d)$ , where a is the first term , d is the common difference and n is the number of terms.
• It is given that sum of n terms of AP is equal to the sum of m terms of the AP.
• Sum of n terms of AP ( $S_n$ ) = $\frac{n}{2}(2a + (n-1)d)$ .
• Sum of m terms of AP ($S_n$ ) = $\frac{m}{2}(2a + (m-1)d)$.

$\frac{n}{2}(2a + (n-1)d)$ = $\frac{m}{2}(2a + (m-1)d)$

$2an + n^{2}d - nd = 2am + m^{2}d - md$

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

$2a(m -n)+d( (m-n)(m+n)) - d(m - n) = 0$

$(m-n)(2a+d( m+n) - d) = 0$

$(m-n)(2a+[( m+n) - 1]d) = 0$

• Therefore, m-n = 0 or (2a+[( m+n) - 1]d) = 0  ..... (Equation 1)

• Now sum of m+n terms of AP

$S_{m+n}$ = $\frac{m+n}{2}[2a + (m+n-1)d]$

But  (2a+[( m+n) - 1]d) = 0  ....(From equation 1)

Substituting the value in above equation ,

$S_{m+n}$ = $\frac{m+n}{2}[0]$

$S_{m+n}$ = 0