Question

first term of an A.P. is a, second term is b and last term is c, then show that sum

of all terms is (a + c)(b +c-2a)

2(b - a) .​

Answer 1

⏯️⏯️⏯️ANSWER⏮️⏮️⏮️

• difference (d)=b-a

• now ...let the nth term is c......
• therefore......

...

$c = a + (n - 1)d \\ c = a + (n - 1)(b - a) \\( c - a) = (n - 1)(b - a) \\ n = 1 + \frac{c - a}{b - a} \\ n = \frac{b + c - 2a}{b - a}$

now the sum of n terms

$s(n) = \frac{n}{2} \times \frac{a + c}{1} \\ = \frac{(b + c - 2a)(a + c)}{2(b - a)}$

Answer 2

Step-by-step explanation:

second term = a+d=b

so d=b-a{eq1}

last term =a+ (n-1)d=c

( let d be the common term by which ap increases or decreases and let us assume ap contains n terms)

c=a + (n-1)(b-a){substitute eq1}

→c-a=(n-1)(b-a)

→(c-a)/(b-a)=n-1

→c-a+b-a/b-a=n{ eq2}

sum up to nterms of an ap

=n/2(2a+(n-1)d)

=n/2(a+c)

=(c+b-2a)(a+c)/2(b-a){sub eq2}