Question

an ap consists of 37 terms the sum of the first three terms of it is 12 and the sum of its last term 3 is 318 then find the first and last terms of the progression

Answer 1

Step-by-step explanation:

n (is number of terms) =37

Sum of first three terms= 12

Sum =[ a+(n-1)d+a+(n-1)d+a+(n-1)d]

12=[ a+(1-1)d+a+(2-1)d+a+(3-1)d]

12=a+0d+a+1d+a+2d

12=3a+3d (divide each term by 3)

4=a+d (1st equation)

Last three terms are 35,36 and 37

Sum of last three terms = 318

Sum =[ a+(n-1)d+a+(n-1)d+a+(n-1)d]

318= [a+(35-1)d+a+(36-1)d+a+(37-1)d]

318=a+34d+a+35d+a+36d

318=3a+105d (divide each term by 3)

106=a+35d (2nd equation)

Subtract 2nd equation from 1st one

a+35d=106

a+d=4

(-) (-) (-)

______

34d=102

d=3

substitute value of d in 1st equation.

a+d=4

a+3=4

a=4-3

a=1

1st term is 1

an=a+(n-1)d

an=1+(1-1)3

an=1

last term

an=a+(n-1)d

an=1+(37-1)3

an=1+(36)(3)

an=1+108

an=109