Question

The sum of 5 th and 9 th terms of an A.P.is 8 and their product is 15. Find the sum of first 28 terms of the A.P.

Answer 1

Solution :

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

The sum of 5th and 9th terms of an A.P. is 8 and their  product is 15.

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

The sum of first 28th term of the A.P.

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

We know that formula of an A.P;

$\boxed{\bf{a_{n}=a+(n-1)d}}}}$

• a is the first term
• d is the common difference.
• n is the term of an A.P.

A/q

$\longrightarrow\sf{a_{5}+a_{9}=8}\\\\\longrightarrow\sf{a+(5-1)d+a+(9-1)d=8}\\\\\longrightarrow\sf{a+4d+a+8d=8}\\\\\longrightarrow\sf{2a+12d=8}\\\\\longrightarrow\sf{2(a+6d)=8}\\\\\longrightarrow\sf{a+6d=\cancel{\dfrac{8}{2} }}\\\\\longrightarrow\sf{a+6d=4..........................(1)}$

&

$\longrightarrow\sf{(a+4d)(a+8d)=15}\\\\\longrightarrow\sf{(4-6d+4d)(4-6d+8d)=15\:\:\:[from(1)]}\\\\\longrightarrow\sf{(4-2d)(4+2d)=15}\\\\\longrightarrow\sf{16\cancel{+8d-8d}-4d^{2} =15}\\\\\longrightarrow\sf{4d^{2} +16=15}\\\\\longrightarrow\sf{-4d^{2} =15-16}\\\\\longrightarrow\sf{\cancel{-}4d^{2} =\cancel{-}1}\\\\\longrightarrow\sf{4d^{2} =1}\\\\\longrightarrow\sf{d^{2} =\dfrac{1}{4} }\\\\\longrightarrow\sf{d=\sqrt{\dfrac{1}{4} } }\\\\\\\longrightarrow\sf{\orange{d=1/2}}$

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

$\longrightarrow\sf{a+5d=4}\\\\\longrightarrow\sf{a+\cancel{6}\times \dfrac{1}{\cancel{2}} =4}\\\\\longrightarrow\sf{a+3=4}\\\\\longrightarrow\sf{a=4-3}\\\\\longrightarrow\sf{\orange{a=1}}$

Now;

We know that formula of the sum of an A.P;

$\boxed{\bf{S_{n}=\frac{n}{2}\big[2a+(n-1)d\big]}}}}$

$\longrightarrow\sf{S_{28}=\cancel{\dfrac{28}{2}} \bigg[2(1)+(28-1)\frac{1}{2} \bigg]}\\\\\\\longrightarrow\sf{S_{28}=14\bigg[2+27\times \dfrac{1}{2} \bigg]}\\\\\\\longrightarrow\sf{S_{28}=14\bigg[2+\dfrac{27}{2} \bigg]}\\\\\\\longrightarrow\sf{S_{28}=14\bigg[\dfrac{4+27}{2} \bigg]}\\\\\\\longrightarrow\sf{S_{28}=\cancel{14}\times \dfrac{31}{\cancel{2}} }\\\\\\\longrightarrow\sf{S_{28}=7\times 31}\\\\\\\longrightarrow\sf{\orange{S_{28}=217}}$

Thus;

The sum of first 28 terms of the A.P. is 217 .

Answer 2

Step-by-step explanation:

a+4d+a+8d=8

⇒2a=8-12d

∴a=4-6d  

(a+4d)(a+8d)=15

⇒(4-6d+4d)(4-6d+8d)=15

⇒(4-2d)(4+2d)=15

⇒16-4d²=15

∴d=1/2

∴a=4-6d=1

S(28)=n/2(a+(n-1)d)=14(1+27/2)=14*29/2=7*29=217