Question

2. In an AP of 50 terms the sum of first 10 terms is 210 and sum of its last 15 terms is 2565,
Find the AP or se terms the sum of fire
Find the AP.​

Answer 1

Answer:

I tried my best to answer your question

Step-by-step explanation:

Your final answer is

AP is -249,-189,-129............ up to 50 terms

I hope u get your answer

thnx for asking

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Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

In an A.P. of 50 terms, the sum of first 10 terms is 210 and sum of it's last 15 terms is 2565.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The Arithmetic progression (A.P.).

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Formula use :

$\bf{\boxed{\bf{Sum\:of\:A.P.=S_{n}=\frac{n}{2} \big[2a+(n-1)d\big]}}}}}$

The sum of first 10 terms is 210 :

$\implies\tt{210=\cancel{\dfrac{10}{2}} \big[2a+(10-1)d\big]}\\\\\\\implies\tt{210=5\big[2a+9d\big]}\\\\\\\implies\tt{\cancel{\dfrac{210}{5} }=2a+9d}\\\\\\\implies\tt{42=2a+9d}\\\\\\\implies\tt{\pink{2a+9d=42..........................(1)}}$

Now;

Sum of the last 15 terms is 2565 :

A/q

$\leadsto\tt{R_{36}\:to\:R_{50}=2565}$

First Term :

$\implies\sf{R_{36}=a+(n-1)d}\\\\\implies\sf{R_{36}=a+(36-1)d}\\\\\implies\sf{R{_{36}=a+35d}}$

Last Term :

$\implies\sf{R_{50}=a+(n-1)d}\\\\\implies\sf{R{_{50}=a+(50-1)d}}\\\\\implies\sf{R_{50}=a+49d}$

So;

$\implies\tt{2565=\dfrac{15}{2} \big[(a+35d)+(a+49d)\big]}\\\\\\\implies\tt{2565=\dfrac{15}{2} \big[a+35d+a+49d\big]}\\\\\\\implies\tt{2565=\dfrac{15}{2} \big[2a+84d\big]}\\\\\\\implies\tt{2565\times 2=15\times 2a+84d}\\\\\\\implies\tt{5130=15\times 2a+84d}\\\\\\\implies\tt{\cancel{\dfrac{5130}{15} }=2a+84d}\\\\\\\implies\tt{342=2a+84d}\\\\\\\implies\tt{\pink{2a+84d=342.......................(2)}}$

Subtracting equation (2) from equation (1),we get;

$\leadsto\sf{\cancel{2a-2a}+9d-84d=42-342}\\\\\leadsto \sf{-75d=-300}\\\\\leadsto \sf{d=\cancel{\dfrac{300}{75} }}\\\\\leadsto \sf{\red{d=4}}$

Putting the value of d in equation (1),we get;

$\leadsto\sf{2a+9(4)=42}\\\\\leadsto\sf{2a+36=42}\\\\\leadsto\sf{2a=42-36}\\\\\leadsto\sf{2a=6}\\\\\leadsto\sf{a=\cancel{\dfrac{6}{2}}} \\\\\leadsto\sf{\red{a=3}}$

Thus;

The A.P. is formed is 3, 7, 11, 15..................