How many terms of the A.P. -6, -frac{11}{2}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the sum of $n$ terms of the given $A.P.$ be $S_n = -25$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Here,the firs

Vedclass · Accepted Answer

$20$ or $5$

Vedclass · Answer

(C) Let the sum of $n$ terms of the given $A.P.$ be $S_n = -25$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Here,the first term $a = -6$ and the common difference $d = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = \frac{1}{2}$.Substituting the values into the formula:$-25 = \frac{n}{2}[2(-6) + (n-1)(\frac{1}{2})]$$-50 = n[-12 + \frac{n}{2} - \frac{1}{2}]$$-50 = n[\frac{n-25}{2}]$$-100 = n^2 - 25n$$n^2 - 25n + 100 = 0$Factoring the quadratic equation:$n^2 - 20n - 5n + 100 = 0$$n(n-20) - 5(n-20) = 0$$(n-20)(n-5) = 0$Thus,$n = 20$ or $n = 5$.Both values are positive integers,so both are valid.

How many terms of the $A.P.$ $-6, -\frac{11}{2}, -5, \ldots$ are needed to give the sum $-25$?

Similar Questions

If $a, b, c, d, e$ are in an arithmetic progression,then what is the value of $a - 4b + 6c - 4d + e$?

If the ratio of the sum of $n$ terms of two arithmetic progressions is $2n + 3 : 6n + 5$,then the ratio of their $13^{th}$ terms is.......

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in

If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.

For $x \geq 0$,the least value of $K$,for which $4^{1+x}+4^{1-x}$,$\frac{K}{2}$,and $16^{x}+16^{-x}$ are three consecutive terms of an $A.P.$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module