Question

$\sf\purple{Maths\: legends\:pls\:help\:!!}$

Show that the sum of all terms of an A.P whose 1st term is a, 2nd term is b and the last term is c is equals to :-

$\sf \dfrac{(a + c)(b + c - 2a)}{2(b - a)}$

Class - 10th

CBSE 2020 (Standard)

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Answer 1

Required Answer:-

GiveN:

• 1st term = a
• 2nd term = b
• last term = c

To prove:

Sum of all terms of the AP is:

$\boxed{\tt \dfrac{(a + c)(b + c - 2a)}{2(b - a)}}$

Proof:

The common difference between two consecutive terms of an AP remains same. That means, common difference = 2nd term - 1st term.

⇒ $\tt{d = b - a}$

Now, by nth term formula,

$\boxed{\tt{a_n = a + (n - 1)d}}$

Last term/nth term given is c. First term is a and d is b - a. Plugging the values to get n,

⇒ $\tt{c = a + (n - 1)b - a}$

⇒ $\tt{(n - 1)b - a = c - a}$

⇒ $\tt{n - 1 = \dfrac{c - a}{b - a} }$

⇒ $\tt{n = \dfrac{c - a}{b - a} + 1}$

That is,

⇒ $\tt{n = \dfrac{c + b - 2a}{b - a} }$

Now, We know the formula for finding the sum of n terms. It is given by:

$\boxed{\tt{S_n = \dfrac{n}{2} \bigg(a + an \bigg) }}$

We have, n = above value we found, a = a and an = c,

⇒ $\tt{S_n = \dfrac{c + b - 2a}{2(b - a)} (a + c)}$

Somewhat, Rearrange to get:

⇒ $\tt{S_n = \dfrac{(a + c)(b + c - 2a)}{2(b - a)} }$

And we are done! :D

Answer 2

GiveN :-

• In an AP , first term is a , 2nd term is b and the last term is c .

To ProvE :-

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

ProoF :-

We know that in an Arthemetic Progression , Common Difference is the difference of two consecutive terms of an AP . Here we are given two consecutive terms that is first term and the second term as a and b respectively .Hence the common difference will be b - a .

Now for finding the sum of n terms of an AP , we need to find which term is n . Here the last term is c , so here let's find out which term is c .

Also we know the formula to find the nth term of an AP as ,

$\sf:\implies\pink{ T_n = 1st\ term + (n-1) d } \\\\\sf:\implies c = a + (n-1) (b-a) \\\\\sf:\implies c - a = (n-1)(b-a) \\\\\sf:\implies \dfrac{c-a}{b-a}= n - 1 \\\\\sf:\implies n = \dfrac{c-a}{b-a} + 1 \\\\\sf:\implies\boxed{\pink{\sf n =\dfrac{c+b-2a}{b-a} }}$

$\rule{200}2$

Now substituting the respective values in the formula to find the sum of n terms of an AP,

$\sf:\implies \pink{ S_n = \dfrac{n}{2}[2a +(n-1)d] }\\\\\sf:\implies S_n =\dfrac{n}{2}[ a + \{ a + (n-1)d\} ] \\\\\sf:\implies S_n =\dfrac{\frac{c+b-2a}{b-a}}{2} [ a + c ] \qquad \bigg\lgroup \red{\tt a+(n-1)d = c}\bigg\rgroup \\\\\sf:\implies S_n =\dfrac{c+b-2a}{2(b-a)} [ a + c]\\\\\sf:\implies \underset{\blue{\sf Hence \ Proved }}{\underbrace{\boxed{\pink{\frak{S_n = \dfrac{(a + c)(b + c - 2a)}{2(b - a)} }}}}}$