Question

(arithmetic progression)
find the following sum :
1+4+7+10+...........+118​

Answer 1

an=1+(n-1)3

118=1+-3+3n

120/3=n

n=40

S40=40/2(1+118)

20(119)

2380

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Answer 2

First term(a)=1

last term(b)=31

common difference(d)=4-1=3

we have,

tn=a+(n-1)d.....(where n=no. of terms)

b=a+(n-1)d

31=1+(n-1)3

.

.

n=11

Thus, 31 is the 11th term.

Now, sum of 11 terms of the given A.S(Sn)

=n/2*(a+b)

=11/2*(1+31)

=11/2*32

=11*16

=176

thus, the sum of n terms, in this case all of them, is 176.

MUNEEPENEE | STUDENT

31+1=32, 28+4=32, 25+7=32...

sum=6*32=192

AJ11 | STUDENT

If you see carefully, you can find that this is the sum of arithmetic sequence by the number of 3

The sum of arithmetic sequence is n/2*(a+L) {n=number of term ,a=first term ,L=last term)

Then.10/2 * ( 1 + 31 )=5 * 32 = 160