The common difference of the A.P. b_{1}, b_{2 | vedclass

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

Question

(B) Let the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$ be $d$.Then the common difference of $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $d

Vedclass · Accepted Answer

$-81$

Vedclass · Answer

(B) Let the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$ be $d$.Then the common difference of $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $d + 2$.For the first $A.P.$,we have $a_{40} = a + 39d = -159$ and $a_{100} = a + 99d = -399$.Subtracting these equations: $(a + 99d) - (a + 39d) = -399 - (-159)$ $\Rightarrow 60d = -240$ $\Rightarrow d = -4$.Substituting $d = -4$ into $a + 39d = -159$: $a + 39(-4) = -159$ $\Rightarrow a - 156 = -159$ $\Rightarrow a = -3$.Now,$a_{70} = a + 69d = -3 + 69(-4) = -3 - 276 = -279$.Given $b_{100} = a_{70}$,we have $b_{100} = -279$.Using the formula for the $n^{th}$ term of the second $A.P.$: $b_{100} = b_{1} + 99(d + 2) = -279$.Substituting $d = -4$: $b_{1} + 99(-4 + 2) = -279 \Rightarrow b_{1} + 99(-2) = -279$.$b_{1} - 198 = -279 \Rightarrow b_{1} = -279 + 198 = -81$.

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$. If $a_{40} = -159$,$a_{100} = -399$ and $b_{100} = a_{70}$,then $b_{1}$ is equal to:

Similar Questions

If the sum of the series $20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + \ldots$ up to $n^{\text{th}}$ term is $488$ and the $n^{\text{th}}$ term is negative,then:

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an arithmetic sequence are $a$,$b$,and $c$ respectively,then the value of $[a(q - r) + b(r - p) + c(p - q)]$ is:

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2, \ldots, x_7$ is $4$ and the mean is $\overline{x}$,then $\overline{x} + x_6$ is equal to:

If $p$ times the $p^{th}$ term of an arithmetic progression is equal to $q$ times its $q^{th}$ term,then the $(p + q)^{th}$ term of this progression is........

If the $p^{th}$ term of an arithmetic progression is $q$ and the $q^{th}$ term is $p$,then its $n^{th}$ term is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module