Question

First second and last term of AP are a,b,2a respectively, its sum is

Answer 1

Heya folk!!

Answer of ur ques is- 3ab / 2(b-a)

A.P : a , b , . . . . . . . . . . . . . .2a

1st term= a1 = a

2nd term = a2= b

nth term = an = 2a

d = a2 - a1 = b-a

an = a1 + (n-1)d = a + (n-1)(b-a) = 2a

(n-1)(b-a) = a

(n-1) = a / (b-a)

n = a / (b-a) + 1 = b / ( b -a )

Sn = n / 2 * ( a1 + an) = b / 2(b-a) * ( a + 2a) = 3ab / 2(b-a)

Hope this helps! Cheers!!

Answer 2

Hi ,

In an A.P

first term = a

second term = b

last term = 2a


common difference ( d ) = b - a

n th term = first term + ( n - 1 ) d = 2a

a + ( n - 1 ) ( b - a ) = 2a

( n - 1 ) ( b - a ) = 2a - a

( n - 1 ) = a / ( b - a )

n = a / ( b - a ) + 1

n = ( a + b - a ) / ( b - a )

n = b / ( b - a ) ---- ( 1 )

sum of the terms ( S )

S = n / 2 [ first term + last term ]

={ b / 2(b - a ) } [ a + 2a ]

= [ b / 2 ( b - a ) ] 3a

S = 3ab / 2(b - a )

I hope this helps you.

:)

=