If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and product of remaining two middle term is $15$, then greatest number of the series will be
If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and product of remaining two middle term is $15$, then greatest number of the series will be
- A
$5$
- B
$7$
- C
$9$
- D
$11$
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- [JEE MAIN 2019]
Let ${a_1},{a_2},\;.\;.\;.\;.,{a_{49}}$ be in $A.P.$ such that $\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,$ then $m = \;\;..\;.\;.\;.\;$
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- [JEE MAIN 2018]
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The sum of the first four terms of an $A.P.$ is $56 .$ The sum of the last four terms is $112.$ If its first term is $11,$ then find the number of terms.
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