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nth Term of A.P

Prove that th...

Question

Prove that the sum of $n$ number of terms of two different A.P.'s can be same for only one value of $n$ if they have same value of $d, a$ .

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Solution

Let $n_{1}$ & $n_{2}$ be no. of terms in two diff AP
sum $\frac{n_{1}}{2} [2 a + (n_{1} - 1) d] = \frac{n_{2}}{2} [2 a + (n_{2} - 1) d]$
$2 a n_{1} + n_{1}^{2} d - n_{1} d = 2 a n_{2} + n_{2}^{1} d - n_{2} d$
$\Rightarrow (n_{1} - n_{2}) [2 a + (n_{1} + n_{2}) d - d] = 0$
$\Rightarrow n_{1} = n_{2}$

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