Let both the series $a_1,a_2,a_3,....$ and $b_1,b_2,b_3,....$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_5=b_9,a_{19}=b_{19}$ and $b_2=0$ , then $a_{11}$ equals

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

Let the first term of the first series be $a_1$ and its common difference be $d_1$. Let the first term of the second series be $b_1$ and its common difference be $d_2$.

Given $a_5 = b_9$, we have $a_1 + 4d_1 = b_1 + 8d_2$. Rearranging gives $a_1 - b_1 = 8d_2 - 4d_1$ (Equation 1).

Given $a_{19} = b_{19}$, we have $a_1 + 18d_1 = b_1 + 18d_2$. Rearranging gives $a_1 - b_1 = 18d_2 - 18d_1$ (Equation 2).

Equating Equation 1 and Equation 2:

$18d_2 - 18d_1 = 8d_2 - 4d_1$

$10d_2 = 14d_1$

$5d_2 = 7d_1$

Since $d_1$ and $d_2$ are prime numbers, it must be that $d_1 = 5$ and $d_2 = 7$.

Given $b_2 = 0$, we have $b_1 + d_2 = 0$, which implies $b_1 = -d_2 = -7$.

Substituting the values of $b_1$, $d_1$, and $d_2$ into Equation 1:

$a_1 = 8d_2 - 4d_1 + b_1 = 8(7) - 4(5) + (-7) = 56 - 20 - 7 = 29$.

Therefore, $a_{11} = a_1 + 10d_1 = 29 + 10(5) = 29 + 50 = 79$.

The value of $a_{11}$ is 79.

Let both the series \(a_1,a_2,a_3,....\) and \(b_1,b_2,b_3,....\) be in arithmetic progression such that the common differences of both the series are prime numbers. If \(a_5=b_9,a_{19}=b_{19}\) and \(b_2=0\) , then \(a_{11}\) equals

The Correct Option is A

Solution and Explanation

Top Questions on Arithmetic Progression

Questions Asked in CAT exam