Question

in an ap sum of first 10 terms is 865 and sum of first14 terms is 799. find a,d,a15​

Answer 1

$\bf\pink{QuestioN:-}$

In an AP sum of first 10 terms is 865 and sum of first 14 terms is 799. Find A, d and A15.

$\bf\purple{AnsweR:-}$

$\Huge\bf\underline{Given:-}$

$\implies\bf{S_{10} = 865}$

$\implies\bf{S_{14} = 799}$

$\Huge\bf\underline{To\:Find:-}$

• A
• d
• A15

$\Huge\bf\underline{Solution:-}$

$\bf{S_n=\dfrac{n}{2}\;[2a+(n-1)\times\:d]}$

According to the given condition, it is already given that,

 

$\implies\bf{S_{10} = 865}$

$\implies\bf{S_{14} = 799}$

 

Then,

 

First case!

$\bf{S_n=\dfrac{n}{2}\;[2a+(n-1)\times\:d]}$

$\bf{S_{10}=\dfrac{10}{2}\;[2a+(10-1)\times\:d]}$

$\bf{S_{10}=5(2a+9d)}$

 

$\bf{\dfrac{865}{5}=2a+9d}$

 

$\bf{173=2a+9d}$ _______ Eq.1

 

Second case!

 

$\bf{S_n=\dfrac{n}{2}\;[2a+(n-1)\times\:d]}$

 

$\bf{S_{14}=\dfrac{14}{2}\;[2a+(14-1)\times\:d]}$

 

$\bf{799=7[2a+13\times\:d]}$

 

$\bf{\dfrac{799}{7}=2a+13d}$

 

$\bf{114.14= 2a+13d}$ _______Eq.2

On subtracting, Eq.1 from Eq.2,

 

$\bf{114.14-173=13d-9d}$

 

$\bf{-58.85=4d}$

 

$\bf{\dfrac{-58.85}{4}=d}$

 

$\bf{d=-14.71}$

 

On substituting the value of d in eq.1,

$\bf{173=2a+9d}$

 

$\bf{173=2a-132.4}$

 

$\bf{173+132.4=2a}$

 

$\bf{305.4=2a}$

 

$\bf{a=\dfrac{305.4}{2}}$

 

$\bf{a= 152.7}$

 

Then,

Now we can find a15.

• A15= A + 14d
• A15 = 152.7 + 14 × -14.7
• A15 = 152- 205.8
• A15 = -53.8

All qstions are solved!

A = 152.7

D = -14.7

A15= -53.8

     

   

     

         

       

     

   

Happy Learning!!♥