Question

if there are n terms in the AP 4,8,12...200 then the sum of n terms

Answer 1

$\underline{\underline{\bold{Question:}}}$

If there are n terms in the AP 4,8,12...200 then the sum of n terms.

$\underline{\bold{Solution:}}$

$\underline{\bold{\textsf{Important formula:}}}\\\\\\\mathbf{1.\:T_n=a+(n-1)d}\\\\\\\mathbf{2.\:S_n=\dfrac{n}{2}[2a+(n-1)d]}\\\\\\\boxed{\mathbf{S_n=\dfrac{n}{2}[a+l]}}$

Here given :

• First term = a = 4.
• Last term = l = 200.
• Difference = d = 4.

$\mathbf{T_n=a+(n-1)d}\\\\\\\implies{\mathbf{200=4+(n-1)4}}\\\\\\\implies{\mathbf{200=4+4n-4}}\\\\\\\implies{\mathbf{4n=200}}\\\\\\\implies{\mathbf{n=\dfrac{200}{4}=50.}}$

No. of terms = 50.

$\mathbf{S_n=\dfrac{n}{2}[a+l]}\\\\\\\implies{\mathbf{S_n=\dfrac{50}{2}[4+200]}}\\\\\\\implies{\bold{S_n=25{\times}204=5100.}}\\\\\\\boxed{\boxed{\mathbf{Sum\:of\;n\:terms\:=S_n=5100.}}}$

Answer 2

Answer:

5100

Step-by-step explanation:

see in Google there will be answer