Question

How many terms of the progression 3,5,7.... must be in order that their sum will be 2600?

Answer 1

Answer :

$\large{\star\:\:\boxed{\sf{S\bf{_{50}}\:=\:2600}}\:\:\star}$

Explanation :

$\dag$ Given :–

• A.P.  :- 3 , 5 , 7 , 9 , ...

• where a = 3 and d = 2 .

$\dag$ To Find :–

• Terms (n) to get a Sum of 2600 (Sₙ)

$\dag$ Formula Applied :–

• $\boxed{\bf{\star\:\:S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] \:\:\star}}$

$\dag$ Solution :–

We have ,

• a = 3
• d = 2
• Sₙ = 2600

Putting these values in the Formula :

$\rightarrow\sf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]}$

$\rightarrow\sf{2600\:=\:\dfrac{n}{2}\:[2(3)\:+\:(n\:-\:1)(2)]}$

$\rightarrow\sf{2600\:=\:\dfrac{n}{2}\:\times\:2[(3)\:+\:(n\:-\:1)]}$

$\rightarrow\sf{2600\:=\:n(3\:+\:n\:-\:1)}$

$\rightarrow\sf{2600\:=\:n(2\:+\:n)}$

$\rightarrow\sf{2600\:=\:2n\:+\:n^2}$

$\rightarrow\sf{n^2\:+\:2n\:-\:2600\:=\:0}$

$\rightarrow\sf{n^2\:+\:52n\:-\:50n\:-\:2600\:=\:0}$

$\rightarrow\sf{n(n\:+\:52)\:-\:50(n\:+52)\:=\:0}$

$\rightarrow\sf{(n\:-\:50)(n\:+52)\:=\:0}$

$\rightarrow\sf{n\:=\:50\:,\:(-52)}$

As we know that number of terms (n) can't be negative . So , we will ignore the negative .

$\rightarrow\boxed{\bf{n\:=\:50}}$

∴ 50 Terms of this A.P. will give a Sum of 2600 .

$\dag$ Additional Information :–

If we were given the Last term (aₙ) then we could find the Number of Terms Easily by Applying the Formula :

$\rightarrow\bf{S_n\:=\:\dfrac{n}{2}\:[a\:+\:a_n]}$