Question

find the sum of first nterm of an A.P 1,4,7,10 also find S40​

Answer 1

Step-by-step explanation:

S40 =2380

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Answer 2

Question :-

find the sum of first nterm of an A.P 1,4,7,10 also find S40.

Answer :-

• $\rm{S_n = \frac{n}{2} [2a + (n - 1)d]}$
• $\rm{S_{40} =2380 }$

Step by step explanation :-

We have,

=> First term (a) = 1

=> Common difference (d) = $\rm a_2- a_1$ = 4 - 1 = 3

=> n term = 40

We have to find,

=> $\rm{S_{40}}$

And we know that,

n term of an A.P. is

$\rm{S_n = \frac{n}{2}[2a + (n - 1)d] }$

Now, find $\rm{S_{40}}$

$\rm:\longmapsto{S_{40} = \frac{40}{2} [2(1) + (40 - 1)3] } \\ \\ \rm:\longmapsto{S_{40} = 20 [2 + (39)3] } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{S_{40} = 20[2 + 117] } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{S_{40} = 20[119)] } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto \red{S_{40} = 2380 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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