Question

if first second and last term of an ap are a,b,2a respectively then find sum of all the terms​

Answer 1

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.

Answer 2

Answer:

Hi,

Let the AP series be x, x+d, x+2d…x+(n-1)d, where;

x is the first term of the sequence;

d is the arithmetic difference;

n is the number of terms.

According to the question;

first term=x=a;……(1)

second term=x+d=b;

d=b-a;……………….(2)

last term=x+(n-1)d=2a;

Substituting the value of d from equation 2 in the above equation, we get;

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

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

n=b/(b-a)…………..(3);

Hence as we know sum of terms of an AP=n/2[2*first term+(n-1)d];

Sum=b/2(b-a)[2a+a/(b-a)(b-a)];

sum=3ab/2(b-a). ANS

Step-by-step explanation:

it can be 100% right