# Calculate the sum 1+4+7+....+31.

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### 6 Answers

Here

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.

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

sum=6*32=192

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

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

sum = 10/2 * ( 1 + 31 )

sum = 5 * 32 = 160

1+(1+3)+(1+2*3)+....+(1+10*3)

=(1+1+...+1)+3*(1+2+3+...+10)=

=11+ (1+10)*30/2=11(1+15)=176

2n-1

n=16

16^2=256

the answer is 256