Find the common difference of an A.P. whose f | vedclass

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

Question

(B) Let the first term be $a = 100$ and the common difference be $d$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(B) Let the first term be $a = 100$ and the common difference be $d$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.The sum of the first six terms is $S_6 = \frac{6}{2}[2(100) + 5d] = 3(200 + 5d) = 600 + 15d$.The sum of the next six terms is the sum of the first $12$ terms minus the sum of the first $6$ terms: $S_{12} - S_6$.$S_{12} = \frac{12}{2}[2(100) + 11d] = 6(200 + 11d) = 1200 + 66d$.Sum of the next six terms $= (1200 + 66d) - (600 + 15d) = 600 + 51d$.According to the problem,$S_6 = 5 	imes (	ext{sum of next six terms})$.$600 + 15d = 5(600 + 51d)$.$600 + 15d = 3000 + 255d$.$15d - 255d = 3000 - 600$.$-240d = 2400$.$d = \frac{2400}{-240} = -10$.

Find the common difference of an $A.P.$ whose first term is $100$ and the sum of whose first six terms is five times the sum of the next six terms.

Similar Questions

The sum of the series $\frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \frac{5}{3! + 4! + 5!} + \dots + \frac{2008}{2006! + 2007! + 2008!}$ is equal to

If the sixth term of a $H.P.$ is $\frac{1}{61}$ and its tenth term is $\frac{1}{105},$ then the first term of that $H.P.$ is

If the $p^{th}$ term of an $A.P.$ is $q$ and the $q^{th}$ term is $p,$ then its $r^{th}$ term is

If $T_n = (n^2 + 1)n!$ and $S_n = T_1 + T_2 + T_3 + ...... + T_n$. Let $\frac{T_{10}}{S_{10}} = \frac{a}{b}$,where $a$ and $b$ are relatively prime natural numbers,then the value of $(b - a)$ is:

The first term of an infinite geometric progression is $x$ and its sum is $5$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module