Question

How many terms of AP -6,-11/2,-5 are needed to give the sum of -25​

Answer 1

Given :-

AP= -6,-11/2,-5 ...

$\sf\ S_n= 25$

To Find

we have to find the number of terms needed to get the sum -25

Solution :-

Using formula

$\bullet\sf\ \ S_n= \dfrac{n}{a}\big\{2a+(n-1)d)\big\}$

We have:-

$\sf\ a= -6 \\ \\ \sf\ d= a_2-a_1\\ \\\sf\ \dfrac{-11}{2}-(-6)\\ \\ \sf\ \ \dfrac{-11+12}{2}\\ \\ \sf\ \ d= \dfrac{1}{2}\\ \\ \sf\ S_n= -25$

Now by putting these values in the formula

$\implies\sf -25= \dfrac{n}{2}\big\{2\times (-6)+(n-1)\times \dfrac{1}{2} \big\}\\ \\ \\ \implies\sf\ -25\times 2= n\bigg(-12+\dfrac{n}{2}-\dfrac{-1}{2}\bigg)\\ \\ \\ \implies\sf\ -50= n\bigg(\dfrac{-24+n-1}{2}\bigg)\\ \\ \\ \implies\sf\ \ -50\times 2=n^2-25n\\ \\ \\ \implies\sf\ n^2-25n+100=0\\ \\ \\ \implies\sf\ n^2-20n-5n+100=0\\ \\ \\ \implies\sf\ n(n-20)-5(n-20)=0\\ \\ \\ \implies\sf\ (n-20)(n-5)=0\\ \\ \\ \therefore\boxed{\purple{\sf \ n= 20\ or\ \ 5}}$