Question

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Answer All Of 2 Question

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Answer 1

Explanation:

Given first term of AP=a

Given second term of AP=b

Given third term of AP=c

Common difference of AP=b−a

Now, a

n

=last term=a+(n−1)d

c=a+(n−1)(b−a)

c−a+b−a=n

______________

b−a

______________

b−a

b+c−2a

=n

Now, sum of AP=S

n

=

2

n

[a+a

n

]

S

n

=

2(b−a)

(b+c−2a)(a+c)

.

Answer 2

Answer:

Sol - 11

Given 1st term of AP=a

Given 2nd term of AP=b

Given 3rd term of AP=c

Common difference of AP=b−a

$now \: a_{n}$ =last term=a+(n−1)d

$\frac{c−a+b−a=n}{ b−a} \\frac{b+c−2a}{b - a} = n$Now, sum of AP$s_{n} = \frac{n}{2} [a+ a_{n} ] \\ s_{n} = \frac{(b+c−2a)(a+c)</p><p>}{2(b - a)}$

Sol -12

For an A.P., let a be the first term and d be the common difference.

The sum of first n terms of an A.P. is given by

Sn = [2a + (n – 1)d]

According to the given condition, Sp = Sq

$⇒  \frac{p}{2} (2a+(p−1)d)= \frac{p}{2} (2a+(q−1)d) \\⇒ p(2a+(p−1)d)=q(2a+(q−1)d) \\ ⇒ 2ap+ {p}^{2} d−pd=2aq+ {q}^{2}d −qd \\ ⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q)=0 \\ ⇒ (p−q)[2a+(p+q)d−d]=0 \\ ⇒ 2a+(p+q)d−d=0 \\ ⇒ 2a+((p+q)−1)d=0 \\⇒ s_{p + q} =0$