Question

The sum of 5th term and 9th term of an AP is 72 and the sum of 7th and 12th terms is 97.
Find the AP.

Answer 1

Answer:

A5 +a9=72

a+4d+a+8d=72

2a+12d=72............eq1

and

a7+a12=97

a+6d+a+11d=97

2a+17d=97.............eq2

subtracting eq1 from eq2

2a+17d=97

2a+12d=72

(-) (-) (-)

__________

5d=25

d=25/5

d=5

put d=5 in eq1

2a+12(5)=72

2a+60 =72

2a=72-60

a=12/2

a=6

Answer 2

Solution :

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

The sum of 5th term and 9th term of an A.P. is 72 and the sum of 7th term and 12th term is 97.

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

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

$\longrightarrow\sf{a_{5}+a_9}=72}\\\\\longrightarrow\sf{a+(5-1)d+a+(9-1)d=72}\\\\\longrightarrow\sf{a+4d+a+8d=72}\\\\\longrightarrow\sf{2a+12d=72}\\\\\longrightarrow\sf{2(a+6d)=72}\\\\\longrightarrow\sf{a+6d=\cancel{\dfrac{72}{2} }}\\\\\longrightarrow\sf{a+6d=36.....................(1)}$

&

$\longrightarrow\sf{a_{7}+a_{12}=97}\\\\\longrightarrow\sf{a+(7-1)d+a+(12-1)d=97}\\\\\longrightarrow\sf{a+6d+a+11d=97}\\\\\longrightarrow\sf{2a+17d=97..........................(2)}$

$\underline{\underline{\bf{Using\:substitution\:method\::}}}}}$

From equation (1),we get;

$\mapsto\sf{a+6d=36}\\\\\mapsto\sf{a=36-6d..........................(3)}$

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

$\mapsto\sf{2(36-6d)+17d=97}\\\\\mapsto\sf{72-12d+17d=97}\\\\\mapsto\sf{72+5d=97}\\\\\mapsto\sf{5d=97-72}\\\\\mapsto\sf{5d=25}\\\\\mapsto\sf{d=\cancel{\dfrac{25}{5} }}\\\\\mapsto\sf{\pink{d=5}}$

Putting the value of d in equation (3),we get;

$\mapsto\sf{a=36-6(5)}\\\\\mapsto\sf{a=36-30}\\\\\mapsto\sf{\pink{a=6}}$

Thus;

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

$\bullet\sf{a=\boxed{6}}}}\\\\\bullet\sf{a+d=6+5=\boxed{11}}}}\\\\\bullet\sf{a+2d=6+2(5)=6+10=\boxed{16}}}\\\\\bullet\sf{a+3d=6+3(5)=6+15=\boxed{21}}}$