Number of terms of an A.P. is even; sum of all odd terms is 24 , sum of all even terms is 30 and las

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8. Sequences and Series

hard

The number of terms of an $A.P.$ is even; the sum of all the odd terms is $24$ , the sum of all the even terms is $30$ and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the $A.P.$ is :

A$4$

B$10$

C$6$

D$8$

(JEE MAIN-2025)

Solution

$a_2+a_4+\ldots+a_n=30$
$a_1+a_3+\ldots+a_{n-1}=24$
$(1)-(2)$
$\left(a_2-a_1\right)+\left(a_4-a_3\right) \ldots\left(a_n-a_{n-1}\right)=6$
$\Rightarrow \frac{n}{2} d=6 \Rightarrow n d=12$
$ a _{ n }- a _1=( n -1) d =\frac{21}{2}$
$\Rightarrow nd – d =\frac{21}{2} \Rightarrow 12-\frac{21}{2}= d$
$\Rightarrow d =\frac{3}{2}, n =8$
$\text { Sum of odd terms }=\frac{4}{2}[2 a+(4-1) 3]=24$
$\Rightarrow a=\frac{3}{2}$
$\text { A.P. } \Rightarrow \frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$
no. of integer terms $=4$

Standard 11

Mathematics

Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.

medium

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easy

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medium

(IIT-1990)

Let $a_{1}, a_{2} \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S _{ n }= a _{1}+ a _{2}+\ldots+ a _{ n }$ If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50,$ then the ordered pair $\left( S _{ n -4}, a _{ n -4}\right)$ is equal to

hard

(JEE MAIN-2020)

If $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ be the $A.M.$ of $a$ and $b$, then $n=$