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Question

If $$1, 2, 3....$$ are first terms; $$1, 3, 5 ....$$ are common differences and $$S_{1}, S_{2}, S_{3} ....$$ are sums of $$n$$ terms of given $$p\ AP's$$; then $$S_{1} + S_{2} + S_{3} + ..... + S_{p}$$ is equal to


A
np(np+1)2
loader
B
n(np+1)2
loader
C
np(p+1)2
loader
D
np(np1)2
loader

Solution

The correct option is A $$\dfrac {np(np + 1)}{2}$$
We know,
$$\sum a=1+2+3+...+p=\dfrac{p(p+1)}{2}$$                                                         ......( 1 )
$$\sum d=1+3+5+...+(2p-1)=\dfrac{p}{2}.[2.1+(p-1)2]=p^2$$                  ......( 2 )
$$S_p=\dfrac{n}{2}.[2a_p+(n-1)d_p]$$                                                                           ......( 3 )

From eq $$(1), (2), (3):$$
$$\therefore S_1+S_2+...+S_p=\dfrac{n}{2}.[2\sum a+(n-1)\sum d]$$

                                   $$=\dfrac{n}{2}.[2\dfrac{p(p+1)}{2}+(n-1).p^2]$$

                                   $$=\dfrac{n}{2}.[p^2+p+np^2-p^2]$$

                                   $$=\dfrac{np(np+1)}{2}$$

Mathematics

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