Question

#Hello Asgardians __/\__

#Solve the following :-

▶️ In a A.P. the sum of first ten terms is -150 and some of next ten terms is -550. Find the A.p...!!!

#Thank you for having a loook :)

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Answer 1

Step-by-step explanation:

Hope it helps u........♥♥

Answer 2

Solution:

Given:

⇒ Sum of ten terms = -150

⇒ Sum of next ten terms = -550

To Find:

⇒ A.P

Formula used:

$\sf{\implies S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

Let the first term of A.P be 'a' and common difference be 'd'.

$\sf{\implies S_{10}=\dfrac{10}{2}[2a+(10-1)d]}$

$\sf{\implies -150=\dfrac{10}{2}[2a+9d]}$

$\sf{\implies -300=20a+90d}$

$\textsf{Take 10 common,}$

$\sf{\implies 2a+9d=-30\;\;\;\;........(1)}$

$\sf{\implies S_{20}=\dfrac{20}{2}[2a+(20-1)d]}$

$\sf{\implies -550-150=\dfrac{20}{2}[2a+19d]}$

$\sf{\implies -700=\dfrac{20}{2}[2a+19d]}$

$\sf{\implies -1400=40a+380d}$

$\textsf{Take 10 common,}$

$\sf{\implies 4a+38d=-140\;\;\;\;........(2)}$

$\sf{Now,\;we\;will\;get\;two\;equation,\;by\;substitution\;method\;we\;solve\;this\;equations,}$

$\sf{\implies 2a+9d=-30\;\;\;\;........(1)}$

$\sf{\implies 2a=-30-9d}$

$\sf{\implies a=\dfrac{-30-9d}{2}}$

$\textsf{Now, put the value of 'a' in Equation (2), we get}$

$\sf{\implies 4a+38d=-140\;\;\;\;........(2)}$

$\sf{\implies 4\Bigg[\dfrac{-30-9d}{2}\Bigg]+38d=-140}$

$\sf{\implies \dfrac{-120-36d}{2}+38d=-140}$

$\sf{\implies \dfrac{-120-36d+76d}{2}=-140}$

$\sf{\implies -120+40d=-280}$

$\sf{\implies 40d=-280+120}$

$\sf{\implies 40d=-160}$

$\sf{\implies d=\dfrac{-160}{40}}$

${\boxed{\sf{\implies d=-4}}}$

$\textsf{Now, put the value of d in Equation (1), we get}$

$\sf{\implies 2a+9d=-30}$

$\sf{\implies 2a+9(-4)=-30}$

$\sf{\implies 2a-36=-30}$

$\sf{\implies 2a=-30+36}$

$\sf{\implies 2a=6}$

$\sf{\implies a=\dfrac{6}{2}}$

${\boxed{\sf{\implies a=3}}}$

$\sf{Thus,\;the\;A.P\;is\;3,\;-1,\;-5...........}$