Class 11

Math

Algebra

Sequences and Series

There are $n$ AM's between $1&31$ such that $7th$ mean: $(n−1)_{th}$ mean $=5:9,$ then find the value of $n.$

**Solution: ****Let the arithmetic mean inserted be $A1,A2,A3,....Am$.**

**now , series is $1,A1,A2,A3,....Am,31$**

**hence , $a=1,T_{n}=31,n=n+2$**

**$T_{n}=a+(n−1)d$**

**$31=1+(n+2−1)d$**

**$30=(n+1)d$**

**$d=n+130 $ _______(1)**

**given that $T_{n}−1T_{7} =95 $**

**$a+(n−1−1+1)da+(7−1+1)d =95 $ **

**(here , $a+(7−1+1)d$ , +1 after 7-1 is because in the series the 1st A.M. is second in terms of series)**

**$a+(n−1)da+7d =95 $**

**$9a+63d=5a+5nd−5d$**

**$4a+68d=5nd$**

**$4a=d(5n−68)$**

**putting value of d frm (1)**

**by solving this you will get value of $n=14$**