Question

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then
find the sum of the first (p + q) terms.
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Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then

find the sum of the first (p + q) terms.

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$\bold{\boxed{ Sn = \frac{n}{2} (2a + (n - 1)d}}$

$= > sp = \frac{p}{2} (2a + (p - 1)d$

$= > sq = \frac{q}{2} (2a + (q - 1)d$

$\bold{\red{sp = sq}}$

$= > \frac{p}{2} (2a + (p - 1)d = \frac{q}{2} (2a + (q - 1)d$

$= > p(2a + (p - 1)d = \frac{2 \times q}{2} (2a + (q - 1)d$

$= > p(2a + (p - 1)d) = q(2a + (q - 1)d)$

$= > 2ap + pd(p - 1) = 2aq + qd(q - 1)$

$= > 2ap - 2aq = qd(q - 1) - pd(p - 1)$

$= > 2a(p - q) = d[(q - 1)q - (p - 1)p]$

$= > 2a(p - q) = d( {q}^{2} - q - ( {p}^{2} - p)$

$= > 2a(p - q) = d( {q}^{2} - q - {p}^{2} + p)$

$= > 2a(p - q) = d( {q}^{2} - {p}^{2} + p - q)$

$= > 2a(p - q) = - d( - {q}^{2} + {p}^{2} - p + q)$

$= > 2a(p - q) = - d( {p}^{2} - {q}^{2} - (p - q))$

$= > 2a(p - q) = - d[(p - q)(p + q) - (p - q)]$

$= > 2a(p - q) = - d[(p - q)(p + q) - (p - q) \times 1]$

Taking (p-q) common from R.H.S

$\bold{= > (p - q)[2a + d(p + q - 1)d] = 0}$

$= > p - q = 0$

$\bold{ = > p = 0}$

$2a + d(p + q - 1) = 0$

Now applying sum formula:

$\bold{ = > sn = \frac{n}{2} [2a + (n - 1)d]}$

$\bold{ = > s(p + q) = \frac{p + q}{2} [2a + (p + q - 1)d]}$

$\bold{= > \frac{p + q}{2} \times 0}$

$= 0$

Therefore,the sum of the first (p + q) terms is 0.

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