Let s_1, s_2, s_3, ldots, s_{10} be the sum o | vedclass

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

Question

(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.For the $k$-th arithmetic progression,th

Vedclass · Accepted Answer

$7260$

Vedclass · Answer

(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.For the $k$-th arithmetic progression,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.Given $n = 12$,the sum $s_k$ is:$s_k = \frac{12}{2} [2(k) + (12-1)(2k-1)]$$s_k = 6 [2k + 11(2k-1)]$$s_k = 6 [2k + 22k - 11] = 6 [24k - 11] = 144k - 66$.Now,we calculate the sum $\sum_{i=1}^{10} s_i$:$\sum_{i=1}^{10} (144i - 66) = 144 \sum_{i=1}^{10} i - \sum_{i=1}^{10} 66$$= 144 	imes \frac{10 	imes 11}{2} - 660$$= 144 	imes 55 - 660$$= 7920 - 660 = 7260$.

Let $s_1, s_2, s_3, \ldots, s_{10}$ be the sum of the first $12$ terms of $10$ arithmetic progressions whose first terms are $1, 2, 3, \ldots, 10$ and whose common differences are $1, 3, 5, \ldots, 19$ respectively. Then $\sum_{i=1}^{10} s_i$ is equal to

Similar Questions

Let $a_1, a_2, \ldots, a_n$ be in an $A$.$P$. If $a_5 = 2a_3$ and $a_{11} = 18$,then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}} + \ldots + \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.

Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of $a^5 b^3 c^2 d$ is $3750 \beta$,then the value of $\beta$ is

If $a_m$ denotes the $m^{th}$ term of an $A.P.$,then $a_m$ =

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

If the $n^{th}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$,then the ratio of their $12^{th}$ terms will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module