Math, asked by 123sona, 11 months ago

solve this problem....

Attachments:

rahimkhan2: hello

123sona: hi

Answers

Answered by Anonymous

$Hey mate Here's an answer below$
Refer to attachment

$<!DOCTYPE html> <html> <head> <title>Dhruv</title> </head> <body> <style> marquee { width: 100%; padding: 10px 0; background-color: lightblue; } </style> <marquee direction="scroll">dhruv$

Attachments:

Answered by siddhartharao77

Answer:

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!

siddhartharao77: 2a + (p + q)d - d? is it a correct equation sis

123sona: ya it's correct

123sona: but I don't understand how we get

123sona: 2a+(p+q)-1]*d=0?

siddhartharao77: Take 'd' as common!

siddhartharao77: (p + q)d - d = 0

It can be written as,

[(p + q) - 1] * d = 0

123sona: oh

123sona: ok

123sona: thank you

siddhartharao77: its ok sis

Previous Question

Next Question