solve this problem....
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rahimkhan2:
hello
hi
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Step-by-step explanation:
Sum of first n term of an AP: s(n) = (n/2)[2a + (n - 1) * d]
(i)
Sum of first p terms = sum of the first q terms
⇒ p/2[2a + (p - 1) * d] = q/2[2a + (q - 1] * d]
⇒ p[2a + pd - d] = q[2a + qd - d]
⇒ 2ap + p²d - pd = 2aq + q²d - qd
⇒ 2ap + p²d - pd - 2aq - q²d + qd = 0
⇒ 2ap - 2aq + p²d - q²d - pd + qd = 0
⇒ 2a(p - q) + (p² - q²)d - d(p - q) = 0
⇒ 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
On multiplying both sides by (p + q)/2, we get
⇒ (p + q/2)[2a + [(p + q) - 1] d] = 0
It is in the form of sum of first (p + q) terms of an AP.
Therefore, sum of first (p + q) terms of an AP = 0.
Hope this helps!
2a + (p + q)d - d? is it a correct equation sis
ya it's correct
but I don't understand how we get
2a+(p+q)-1]*d=0?
Take 'd' as common!
(p + q)d - d = 0
It can be written as,
[(p + q) - 1] * d = 0
It can be written as,
[(p + q) - 1] * d = 0
oh
ok
thank you
its ok sis
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