Insert five numbers between $8$ and $26$ such that resulting sequence is an $A.P.$
Insert five numbers between $8$ and $26$ such that resulting sequence is an $A.P.$
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, $a=8, b=26, n=7$
Therefore, $26=8+(7-1) d$
$\Rightarrow 6 d=26-8=18$
$\Rightarrow d=3$
$A_{1}=a+d=8+3=11$
$A_{2}=a+2 d=8+2 \times 3=8+6=14$
$A_{3}=a+3 d=8+3 \times 3=8+9=17$
$A_{4}=a+4 d=8+4 \times 3=8+12=20$
$A_{5}=a+5 d=8+5 \times 3=8+15=23$
Thus, the required five numbers between $8$ and $26$ are $11,14,17,20$ and $23 .$
Similar Questions
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
- [JEE MAIN 2017]
If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in $H.P.$, then
If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in $H.P.$, then
If the sum of the first $n$ terms of a series be $5{n^2} + 2n$, then its second term is
If the sum of the first $n$ terms of a series be $5{n^2} + 2n$, then its second term is
The sum of $n$ arithmetic means between $a$ and $b$, is
The sum of $n$ arithmetic means between $a$ and $b$, is
Let ${T_r}$ be the ${r^{th}}$ term of an $A.P.$ for $r = 1,\;2,\;3,....$. If for some positive integers $m,\;n$ we have ${T_m} = \frac{1}{n}$ and ${T_n} = \frac{1}{m}$, then ${T_{mn}}$ equals
Let ${T_r}$ be the ${r^{th}}$ term of an $A.P.$ for $r = 1,\;2,\;3,....$. If for some positive integers $m,\;n$ we have ${T_m} = \frac{1}{n}$ and ${T_n} = \frac{1}{m}$, then ${T_{mn}}$ equals
- [IIT 1998]