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

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ayush01's profile pic

Posted on

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.

aj11's profile pic

Posted on

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

 

 

fahad-alfahad's profile pic

Posted on

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

sum = 10/2 * ( 1 + 31 )

sum = 5 * 32 = 160

giorgiana1976's profile pic

Posted on

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

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