tu tne 1unbwing questions. (Any one)14 MarksIf first term of an A.P. i

1.	tu tne 1unbwing questions. (Any one)14 MarksIf first term of an A.P. is a, second term is b and last term is c, then show that sum of allthe terms is (a + C) (b +c-2a(atc) (b+c- 2a2 (b-a).
Answer» ANS : AP :a , b ............ ,c First term (a₁)= a Common Difference (d)= (b - a) Last term ie, nth term (aₙ) = c Apply the Formula:aₙ= a₁+ (n-1)d So :c=a+ (n-1) [b-a] c - a = (n-1) [b-a] (n-1)= (c-a) / [b-a] => n ={(c-a) / [b-a] } +1 ={ (c-a) + (b-a) } / [b-a] = {b+c - 2a} / [b-a] Therefore :n ={b+c - 2a} / [b-a]-------(1) To determine Sum of n term (Sn) =>Apply :Sₙ = [n / 2 ] {a₁+aₙ} Sn = [n / 2 ] * {a+c} -----------(2) From (1) :n ={b+c - 2a} / [b-a], Substitute in(2) . Sₙ = [ {n}/ 2 ] * {a+c} Sₙ = [{{b+c - 2a} / [b-a]}* (1 /2) ] * {a+c} Sₙ=[(a+c) (b+c - 2a) ] / 2 (b-a) Hence Sum of n terms : Sₙ =[(a+c) (b+c - 2a) ] / 2 (b-a) (HENCE PROVED) IF YOU ARE SATISFIED DO GIVE THUMBS UP.

tu tne 1unbwing questions. (Any one)14 MarksIf first term of an A.P. is a, second term is b and last term is c, then show that sum of allthe terms is (a + C) (b +c-2a(atc) (b+c- 2a2 (b-a).

Answer»

ANS : AP :a , b ............ ,c

First term (a₁)= a

Common Difference (d)= (b - a)

Last term ie, nth term (aₙ) = c

Apply the Formula:aₙ= a₁+ (n-1)*d

So :c=a+ (n-1) *[b-a]

c - a = (n-1) *[b-a]

(n-1)= (c-a) / [b-a]

=> n ={(c-a) / [b-a] } +1 ={ (c-a) + (b-a) } / [b-a] = {b+c - 2a} / [b-a]

Therefore :n ={b+c - 2a} / [b-a]-------(1)

To determine Sum of n term (Sn) =>Apply :Sₙ = [n / 2 ] * {a₁+aₙ}

Sn = [n / 2 ] * {a+c} -----------(2)

From (1) :n ={b+c - 2a} / [b-a], Substitute in(2) .

Sₙ = [ {n}/ 2 ] * {a+c}

Sₙ = [{{b+c - 2a} / [b-a]}* (1 /2) ] * {a+c}

Sₙ=[(a+c) *(b+c - 2a) ] / 2 *(b-a)

Hence Sum of n terms : Sₙ =[(a+c) *(b+c - 2a) ] / 2 *(b-a)

(HENCE PROVED)

IF YOU ARE SATISFIED DO GIVE THUMBS UP.