Insert five numbers between 8 and 26 such tha | vedclass

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

Question

(A) Let $A_{1}, A_{2}, A_{3}, A_{4},$ and $A_{5}$ be five numbers between $8$ and $26$ such that $8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26$ is an $A.P

Vedclass · Accepted Answer

$11, 14, 17, 20, 23$

Vedclass · Answer

(A) Let $A_{1}, A_{2}, A_{3}, A_{4},$ and $A_{5}$ be five numbers between $8$ and $26$ such that $8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26$ is an $A.P.$Here,the first term $a = 8$ and the last term $l = 26$. The total number of terms $n = 5 + 2 = 7$.The formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a + (n - 1)d$.Substituting the values: $26 = 8 + (7 - 1)d$.$26 = 8 + 6d$ $\Rightarrow 6d = 18$ $\Rightarrow d = 3$.The five numbers are:$A_{1} = a + d = 8 + 3 = 11$$A_{2} = a + 2d = 8 + 6 = 14$$A_{3} = a + 3d = 8 + 9 = 17$$A_{4} = a + 4d = 8 + 12 = 20$$A_{5} = a + 5d = 8 + 15 = 23$Thus,the required five numbers are $11, 14, 17, 20, 23$.

Insert five numbers between $8$ and $26$ such that the resulting sequence is an $A.P.$

Similar Questions

What is the arithmetic mean of the series $a + (a + d) + (a + 2d) + \dots + (a + 2nd)$?

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

If $a$,$b$,and $c$ are positive real numbers,then $a/b + b/c + c/a$ is greater than or equal to what value?

The sum of all two-digit positive numbers which,when divided by $7$,yield $2$ or $5$ as a remainder is:

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,then find the number of terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module