Question

The first , second and the last terms of an A.P are a, b , and c respectively. Prove that their sum is (a+c)(b+c -2a) / 2(b - a)...

It's a 4 mark Question so answer it with full explanation..​

Answer 1

Given : First term of the AP is 'a' and let the common difference of the AP be 'd'

According to the question :

Second term of the AP = b

⇒ a + ( 2 - 1 )d = b

⇒ a + d = b

⇒ d = b - a → ( 1 )

Last term of the AP ( l ) = aₙ = c

⇒ a + ( n - 1 )d = c

⇒ ( n - 1 )d = c - a

From Eq( 1 )

⇒ ( n - 1 )( b - a ) = c - a

⇒ n - 1 = ( c - a ) / ( b - a )

⇒ n = ( c - a )/( b -a ) + 1

⇒ n = ( c - a + b - a ) / ( b - a )

⇒ n = ( b + c - 2a ) / ( b - a ) → ( 2 )

Now let's find the sum of AP

Sum of n terms of an AP ( Sₙ ) = n/2 × ( a + l )

⇒ Sₙ = n/2 × ( a + c )

From Eq( 2 )

$\Rightarrow \sf S_n=\dfrac{\dfrac{b+c-2a}{b-a} }{2} \times ( a + c ) \\\\\\ \Rightarrow \sf S_n=\dfrac{b+c-2a}{2(b-a)} \times ( a+c) \\\\\\ \Rightarrow \sf S_n=\dfrac{(a+c)(b+c-2a)}{2(b-a)}$

Hence Proved.

Answer 2

$\bf{\blue{\bigstar}}$ GiveN :

• The first , second and last terms of AP are a , b and c respectively.

$\bf{\blue{\bigstar}}$ To ProvE :

• $\bold{\frac{(a+c)(b+c-2a)}{2(b-a)} }$

$\bf{\blue{\bigstar}}$ ProoF :

General term of an AP = $\bold{a_n=a+(n-1)d}$

Given , First term  = a

Second term , a + ( 2 - 1 ) d = b

⇒ a + d =b

⇒ d = b - a ... (1)

Last term , a + ( n - 1 ) d = c

⇒ ( n - 1 ) d = c - a

⇒ ( n - 1 ) ( b - a ) = c - a [ From (1) ]

⇒ n - 1 = ( c - a ) / ( b - a )

⇒ n = (c - a ) / ( b - a ) + 1

⇒ n = ( c + b - 2a ) / ( b - a ) ... (2)

Now ,Sum of n terms in AP ,

• $\bold{S_n=\frac{n}{2}(a+l) }$

where ,

• n = nth term
• a = First term
• l = last term

$\bold{S_n=\frac{n}{2}(a+l) }\\\\\implies \bold{S_n=\frac{\frac{(c+b-2a)}{b-a}}{2} (a+c) \;\;[From\;(2)]}\\\\\implies \bold{\bf{\blue{S_n=\frac{(a+c)(b+c-2a)}{2(b-a)} }}}$