Insert five numbers between 11 and 29 such that the resulting sequence is an AP.

Question:

Insert five numbers between 11 and 29 such that the resulting sequence is an AP.

 

Solution:

To find: Five numbers between 11 and 29, which are in A.P.

Given: (i) The numbers are 11 and 29

Formula used: (i) $A_{n}=a+(n-1) d$

Let the five numbers be $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$

According to question $11, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ and 29 are in A.P.

We can see that the number of terms in this series is 7

For the above series:-

$a=11, n=7$

$\mathrm{A}_{7}=29$

Using formula, $A_{n}=a+(n-1) d$

$\Rightarrow \mathrm{A}_{7}=11+(7-1) \mathrm{d}=29$

$\Rightarrow 6 \mathrm{~d}=29-11$

$\Rightarrow 6 \mathrm{~d}=18$

$\Rightarrow \mathrm{d}=3$

We can see from the definition that $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$ are five arithmetic mean between 11 and 29 , where $d=3, a=11$

Therefore, Using formula of arithmetic mean i.e. $A_{n}=a+n d$

$A_{1}=a+n d$

$=11+3$

$=14$

$\mathrm{A}_{2}=\mathrm{a}+\mathrm{nd}$

$=11+(2) 3$

$=17$

$\mathrm{~A}_{3}=\mathrm{a}+\mathrm{nd}$

$=11+(3) 3$

$=20$

$A_{4}=a+n d$

$=11+(4) 3$

$=23$

$A_{5}=a+n d$

$=11+(5) 3$

$=26$

Ans) 14, 17, 20, 23 and 26 are the required numbers. 

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