Question

If the sum of p terms of AP is q and the sum of q terms is p, what will the sum of p+q terms be?​

Answer 1

Answer:

It will be zero.

Step-by-step explanation:

tp=a+(p-1)d

tq=a+(q-1)d

According to given condition

tp=q

tq=p

Sp+q=(p+q)÷2×(a+(p+q-1)d)

Supstituting the values of p and q

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

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$\star$ SOME Useful FORMULAS:-

FOr an A.P. series with first term a and last term l and common difference (d).....

• n'th term is

$\bf\implies \boxed{\bf\red{a_n=a+(n-1)d}}$

•Sum of n'th terms

$\bf\implies \boxed{\bf\red{S_n=\frac{n[2a+(n-1)d]}{2}}}$

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$\bf\star$GIVeN :-

•condition:-1

sum of p terms of AP is q ..

$\bf\implies S_p=\frac{p[2a+(p-1)d]}{2}=q$

•condition:-1

sum of p terms of AP is q ..

$\bf\implies S_q=\frac{q[2a+(q-1)d]}{2}=p$

$\bf\star$To Find:-

$\bf\implies S_{p+q}=?$

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Now....

from (i) ,,,

$\bf\implies \frac{p[2a+(p-1)d]}{2}=q\\ [tex]\bf\implies p[2a+(p-1)d]=2q\\ \bf\implies 2a+(p-1)d=\frac{2q}{p}\\ \bf\implies 2a=(p-1)d+\frac{2q}{p}.......(iii)$

Now from (ii)

$\bf\implies \frac{q[2a+(q-1)d]}{2}=p\\ [tex]\bf\implies q[2a+(q-1)d]=2p\\ \bf\implies 2a+(p-1)d=\frac{2p}{q}\\ \bf\implies 2a=(q-1)d+\frac{2p}{q}..........(iv)$

Now comparing equation (iii) and (iv)

$\bf\implies (p-1)d+\frac{2q}{p}=(q-1)d+\frac{2p}{q}\\ \bf\implies d((p-1)-(q-1))=\frac{2p}{q}-\frac{2q}{p}\\ \bf\implies d(p-\cancel1-q+\cancel1)=\frac{2p{}^{2}-2q{}^{2}}{pq}\\ \bf\implies d(p-q)=\frac{2(p{}^{2}-q{}^{2})}{pq}\\ \bf\implies d=\frac{2(p+q)(p-q)}{pq}\\ \bf\implies d=\frac{2(p+q)}{pq}$

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$\bf\implies S_{p+q}=\frac{(p+q)(2a+(p+q-1)d}{2}\\ \bf\implies S_{p+q}= \frac{(p(2a+(p+q-1)d)}{2}+\frac{(q(2a+(p+q-1)d)}{2}\\ \bf\implies S_{p+q}= \frac{p(2a+(p-1)d+qd)}{2}+\frac{(q(2a+(q-1)d)+pd}{2}\\ \bf\implies S_{p+q}= \frac{p(2a+(p-1)d)}{2}+\frac{qd}{2}+\frac{q(2a+(q-1)d}{2}+\frac{pd}{2}\\ \bf\implies S_{p+q}=S_p+\frac{qd}{2}+S_q+\frac{pd}{2}\\ \bf\implies S_{p+q}=q+\frac{qd}{2}+p+\frac{pd}{2}\\ \bf\implies S_{p+q}=(p+q)+\frac{d}{2}(p+q)$

$\bf\implies S_{p+q}=(p+q)+\frac{\frac{\cancel{2}(p+q)}{pq}}{\cancel{2}}(p+q)\\ \bf\implies S_{p+q}=(p+q)+\frac{(p+q){}^{2}}{pq}\\ \bf\implies \boxed{\bf\red{S_{p+q}=\frac{(p+q)}{pq}(p+q+pq)}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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