Question

The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times its 18th term, find the A.P.

Answer 1

Solution : (Ques. Error )

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

The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times it's 8 term.

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

The A.P.

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

We know that formula of an A.P;

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

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

A/q

$\longrightarrow\sf{a+(5-1)d+a+(9-1)d=30}\\\\\longrightarrow\sf{a+4d+a+8d=30}\\\\\longrightarrow\sf{2a+12d=30}\\\\\longrightarrow\sf{2(a+6d)=30}\\\\\longrightarrow\sf{a+6d=\cancel{\dfrac{30}{2} }}\\\\\longrightarrow\sf{a+6d=15.......................(1)}$

&

$\longrightarrow\sf{a_{25}=3\times a_{8}}\\\\\longrightarrow\sf{a+(25-1)d=3\times [a+(8-1)d]}\\\\\longrightarrow\sf{a+24d=3\times (a+7d)}\\\\\longrightarrow\sf{a+24d=3a+21d}\\\\\longrightarrow\sf{a-3a=21d-24d}\\\\\longrightarrow\sf{\cancel{-}2a=\cancel{-}3d}\\\\\longrightarrow\sf{2a=3d}\\\\\longrightarrow\sf{a=\dfrac{3d}{2}....................(2) }$

Now;

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

$\longrightarrow\sf{\dfrac{3d}{2} +6d=15}\\\\\\\longrightarrow\sf{3d+12d=30}\\\\\\\longrightarrow\sf{15d=30}\\\\\\\longrightarrow\sf{d=\cancel{\dfrac{30}{15} }}\\\\\\\longrightarrow\sf{\orange{d=2}}$

Putting the value of d in equation (2),we get;

$\longrightarrow\sf{a=\dfrac{3(2)}{2} }\\\\\\\longrightarrow\sf{a=\cancel{\dfrac{6}{2}} }\\\\\\\longrightarrow\sf{\orange{a=3}}$

Thus;

$\boxed{\bf{Arithmetic\:progression\::}}}$

$\bullet\:\sf{a=\boxed{3}}}\\\\\bullet\:\sf{a+d=3+2=\boxed{5}}}\\\\\bullet\:\sf{a+2d=3+2(2)=3+4=\boxed{7}}}\\\\\bullet\:\sf{a+3d=3+3(2)=3+6=\boxed{9}}}$

Answer 2

$\mathbb{\huge{\underline{\underline{\red{Question:-}}}}}$

✒ The sum of the 5th term and the 9th term of an A.P. is 30. If its 25th term is three times its 18th term, find the A.P.

$\mathcal{\huge{\underline{\underline{\orange{Answer:-}}}}}$

⭐ The Following AP For the Question is 3, 5, 7, 9, 11, 13, 15, 19,...............

$\mathcal{\huge{\underline{\underline{\green{Solution:-}}}}}$

Given :-

The sum of the 5th term and the 9th term of an A.P. is 30.

To Find :-

The AP if its 25th term is three times its 18th term.

☑ Calculation :-

The sum of 5th and 9th term is 30.

⇒ a + 4d + a + 8d = 30

⇒ 2a + 12d = 30 ________(1)

And,

Its 25th term is 3 times its 8th term.

⇒ a + 24d = 3(a + 7d)

⇒ a + 24d = 3a + 21d

⇒ a + 24d - 3a - 21d = 0

⇒ - 2a + 3d = 0 _________(2)

Adding equation (1) and (2).

⇒ 2a +12d = 30 

  - 2a + 3d =   0

________________

           15d = 30

________________

⇒ 15d = 30

⇒ d = 30/15

⇒ d = 2

So, common difference, 'd' is 2.

Putting the value of d = 2 in the equation (1).

2a + 12d = 30

⇒ 2a + (12*2) = 30

⇒ 2a = 24 = 30

⇒ 2a = 30 - 24

⇒ 2a = 6

⇒ a = 6/2

⇒ a = 3

So, the first term of the required AP is 3

Hence , the required AP is 3, 5, 7, 9, 11, 13, 15, 19,.............

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