Question

Pls help with an elaborate answer

Answer 1

Answer:

By definition,

the nth term is going to be the difference of the sum of the first n-1 terms from the first n terms

Or mathematically

Tn = Sn - Sn-1

Tn = (2n^2-3n)-(2(n-1)^2 - 3(n-1))

= 2n^2 - 3n - (2n^2 -7n+5)

= 4n-5

So, the general nth term is gonna be 4n-5.

Hope this helps you !

Answer 2

$</p><p>\huge{\fbox{\fbox{\rightarrow{\mathfrak {\green {Answer \: - \: OPTION \: C}}}}}}$

$</p><p>\underline {\underline {\orange {Step-By-Step-Explaination \: - }}}$

$</p><p>\huge {\boxed {\boxed {\bold{\underline {\mathfrak{\blue{Question\: : }}}}}}}$

WHAT IS THE NTH TERM OF AN ARITHMETIC PROGRESSION WHOSE SUM TO 'N' TERMS IS [ 2n^2 - 3n ] ?

$</p><p>\huge {\boxed {\boxed {\bold{\underline {\mathfrak{\blue{Options\: : }}}}}}}$

[A] 4n - 3

[B] 4n - 5

[C] 4n + 5

[D] 5 - 4n

$</p><p>\huge {\boxed {\boxed<strong> </strong>{\bold{\underline {\red{Solution\: : }}}}}}$

We know that in an A.P of n terms, the sum is

$\frac{n}{2} (2a + (n - 1)d)$

This is equal to 2n^2 - 3n

$\frac{n}{2} (2a + (n - 1)d) = 2 {n}^{2} - 3n$

Sum upto n terms is 2n^2 - 3n.

Sum upto (n-1) terms is 2(n-1)^2 - 3(n-1).

Hence, nth term = Sn - Sn-1

Tn $= 2 {n}^{2} - 3n - 2 {(n - 1)}^{2} + 3(n - 1)$

$4n - 5$

Hence Option C is correct.