If a1-a2-a3-an are consecutive terms of an increasing A-P- and 12-a1-22-a2-32-a3-n2-an-n-1nn-13- then the value of a5-a3-a26 is equal to

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Arithmetic Progression

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3}, . . . . . . . . . ., a_{n}$ are consecutive terms of an increasing $A . P .$ and $(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3}$ , then the value of $(\frac{a_{5} + a_{3} - a_{2}}{6})$ is equal to

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Solution

The correct option is A $1$
Given, $a_{1}, a_{2} . . . a_{n}$ are in A.P
Also, $(1^{2} - a_{1}) + (2^{2} - a_{2}) + . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3} . . . (1)$
$\Rightarrow (1^{2} + 2^{2} + . . . n^{2}) - (a_{1} + a_{2} + . . . + a_{n}) = \frac{(n - 1) n (n + 1)}{3}$
$\sum n^{2} - \frac{n}{2} (a_{1} + a_{n}) = \frac{n (n - 1) (n + 1)}{3}$ [ sum of AP]
$\frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n - 1) (n + 1)}{3} = \frac{n}{2} (a_{1} + a_{n}) [\sum n^{2}$ formula]
$\frac{n + 1}{3} [\frac{2 n + 1}{2} - (n - 1)] = \frac{a_{1} + a_{n}}{2}$
$n + 1 = a_{1} + a_{n} . . . (2)$
putting n =1 in (1) $\Rightarrow 1^{2} - a_{1} = 0 \Rightarrow a_{1} = 1$
from(2), $a_{n} = n$
$∴ (\frac{a_{5} + a_{3} - a_{2}}{6}) = (\frac{a_{5} + d}{6}) = \frac{a_{6}}{6} = \frac{6}{6} = 1$ [ $∵ a_{3} - a_{2} = d]$ common diff.
option A is correct

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}

are first n terms of an A.P. with first term

a_{1} = 0

and common difference

d \neq 0

Then value of

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

Q. If for an

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

and

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

equals

Q. If

a_{r} > 0, r \in N

and

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

is equal to

For an increasing A.P. $a_{1}, a_{2}, a_{3} . . . a_{n}$ ,
if $a_{1} + a_{3} + a_{5} = - 12$ and $a_{1} \cdot a_{3} \cdot a_{5} = 80$ , then which of the following is true.

Q. If

a_{r} > 0; \forall r, n \in N

and

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

are in A.P, then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

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If a1,a2,a3,..........,an are consecutive terms of an increasing A.P. and (12−a1)+(22−a2)+(32−a3)+.......+(n2−an)=(n−1)n(n+1)3, then the value of (a5+a3−a26) is equal to

If $a_{1}, a_{2}, a_{3}, . . . . . . . . . ., a_{n}$ are consecutive terms of an increasing $A . P .$ and $(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3}$ , then the value of $(\frac{a_{5} + a_{3} - a_{2}}{6})$ is equal to