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If 1, 2, 3,..... are first terms, 1, 3, 5,...... are common di

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 AP's, then $S_{1} + S_{2} + S_{3} + ... + S_{p}$ is equal to

\frac{n p ( n p + 1 )}{2}

\frac{n ( n p + 1 )}{2}

\frac{n p ( p + 1 )}{2}

\frac{n p ( n p - 1 )}{2}

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Jack Discussed

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 AP's, then

S_{1} + S_{2} + S_{3} + ... + S_{p}

is equal to

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Text solutionVerified

Let

a_{1} = 1, a_{2} = 2, a_{3} = 3, ....., a_{p} = p

be the first terms of APs

and

d_{1} = 1, d_{2} = 3, d_{3} = 5, ......, d_{p} = 2 p - 1;

be the common differences of the respective AP

S_{1} + S_{2} + S_{3} + .... S_{p} = \frac{n}{2} (2 a_{1} + (n - 1) d_{1}) + \frac{n}{2} (2 a_{2} + (n - 1) d_{2}) + .... + \frac{n}{2} (2 a_{p} + (n - 1) d_{p}) = n (a_{1} + a_{2} + a_{3} + .... + a_{p}) + \frac{n ( n - 1 )}{2} (d_{1} + d_{2} + d_{3} + .... + d_{p}) = n \frac{p ( p + 1 )}{2} + \frac{n ( n - 1 )}{2} p^{2} = \frac{n p}{2} (p + 1 - p n - p) = \frac{n p}{2} (p n + 1)

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Question Text	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 AP's, then $S_{1} + S_{2} + S_{3} + ... + S_{p}$ is equal to
Answer Type	Text solution:1
Upvotes	150