Question

Find the sum of first n terms of an AP whose nth term is (5n -1).
Hence, find the sum of first 20 terms.​

Answer 1

Answer:

The nth term of the ap is given as 5n - 1.

Therefore the first term of the ap (a) = 5(1) - 1 = 4

And the second term will be = 5(2) - 1 = 9

The common difference now will be (d) = 9 - 4 = 5

The sum of n terms of an ap =

$\frac{n}{2} (2a + (n - 1)d)$= n/2( 2(4) + (n - 1)5) = n/2( 3 + 5n)

Sum of first 20 term terms = n/2( 3 + 5n ) = 10( 3 + 100) = 1030.

[P.S.- Bare in mind that we were told to first find the sum of first n term of an ap then then using that we were to find the sum of the first 20 terms... that's the reason why I used Sn = n/2( 3 + 5n ) and not Sn = n/2( 2a + (n - 1)d)]

Answer 2

${\huge{\underline{\underline{\rm{Question:}}}}}$

Find the sum of first n terms of an A.P. whose n th term is (5n - 1) , also find the sum of first 20 terms .

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

★ Given :

• n th term = (5n - 1)

★ To Find :

• Sum of n terms .
• Sum of first 20 terms .

★ Solution :

# Finding the first term (a) and common difference (d) .

Put , n = 1

so ,

$\mapsto \sf \: first \: term \: = (5 \times 1 - 1)$

$\boxed{\sf \: first \: term(a_{1} ) = 5 - 1 = 4}$

Put , n = 2

to find second term

so ,

$\mapsto \sf \: second \: term = (5 \times 2 - 1)$

$\boxed{ \sf \: second \: term(a_{2} ) = 10 - 1 = 9}$

Now ,

$\sf \: common \: difference =a _{2} - a_{1}$

$\sf \: common \: difference = 9 - 4$

$\boxed{ \sf \: common \: difference = 5}$

Now , we have got the value of a and d

So , put it in the Formula for finding Sum

$\rm \star \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$

$\rm \mapsto \: s_{n} = \dfrac{n}{2} (2×4 + (n - 1)5)$

$\mapsto \rm \: s_{n} = \dfrac{n}{2} (8 + 5n - 5)$

$\mapsto \rm \: s_{n} = \dfrac{n}{2} (5n +3)$

$\large\mapsto \boxed{\rm \: s _{n} = \dfrac{5 {n}^{2} +3n }{2} }$

Now ,

We have to find the 20th term ,

so ,

Put n=20 in the Formula of Sn

$\mapsto \rm \: s_{20} = \dfrac{5 \times {(20)}^{2} +3× 20 }{2}$

$\mapsto \rm \: s_{20} = \dfrac{5 \times 400 +60}{2}$

$\mapsto \rm \: s_{20} = \dfrac{ 2000 + 60}{2}$

$\mapsto \rm \: s _{20} = \dfrac{ \cancel{2060}}{ \cancel2}$

$\large \boxed{ \rm \: s_{20} = 1030}$

So , we have got both the answers ; they are :

1. $\tt \: \dfrac{5 {n}^{2} +3 n}{2}$
2. $\tt \: 1030$