Question

Find AP whose sum and product of 3 adjacent terms are 39 and 1560 respectively nth term of a sequence is given by the formula an = 10-3n ,find the sum of its 20 terms .​

Answer 1

Step-by-step explanation:

Given:

Sum of 3 adjacent terms=39

Product of 3 adjacent terms=1560

an=10-3n

To find:

A.P formed

Sum of its 20 terms

solution:

Let 3 adjacent terms be

a-d, a, a+d

sum=a-d+a+a+d=3a

but it is given that sum = 39

therefore,

3a=39

a=13 - - - - 1

Product= (a-d)(a)(a+d)

=(a^2-ad)(a+d)

=a^3+a^2d-a^2d-ad^2

=a^3-ad^2

but it is given that product = 1560

therefore,

a^3-ad^2=1560

a(a^2-d^2)=1560- - - - 2

from 1 a= 13

putting in 2, we get,

13(13^2-d^2)=1560

169-d^2=1560/13

-d^2=1560/13-169

-d^2=1560/13-2197/13

-d^2=-637/13

d^2=49

d=7

Now A.P formed:

=a,a+d,a+2d- - -

=13,20,27- - -

sum of its 20 terms=n/2(2a+(n-1)d)

here

n=20 (given)

a=13(find above)

d=7(find above)

putting values we get

=20/2(2*13+(20-1)7)

=10(26+133)

=10*159

=1590