Question

If the 10th term of an A.P is 52 and 17th term is 20 more than the 13th term, find A.P. ​

Answer 1

${\huge{\underline{\underline{\sf{\blue{SOLUTION:-}}}}}}$

Question

⠀⠀⠀⠀ • If the 10th term of an A.P is 52 and 17th ⠀⠀⠀⠀term is 20 more than the 13th term, find A.P.

Given

⠀⠀⠀⠀ • a10 = 52

⠀⠀⠀⠀ • a17 = 20+a13

To Calculate

⠀⠀⠀⠀ • Find A.P

Step by step explanation

⠀⠀

⠀⠀⠀⠀ ⠀ Using formula ⤵️

⠀⠀

⠀⠀⠀ ⠀ ⠀⠀ ☆ an = a+(n-1)×d ☆

⠀⠀

• a10 = 52 [ Given]

Putting values,

➡️ 52= a+(10-1)×d

➡️ 52 = a+9d________________(1)

• a17 = 20+a13 [ Given]

Putting values,

➡️ a+16d= 20+a+12d

➡️ 16d-12d = 20

➡️ 4d= 20

➡️ d= 20/4

➡️ d= 5

Putting value of d in eq 1

• 52= a+9d

➡️ 52= a+9×5

➡️ 52= a+45

➡️ a= 52-45

➡️ a= 7

⠀⠀

So, The A.P is 7, 12, 17,22.........

______________________________________________

Answer 2

$\bf{\underline{\underline{Question:-}}}$

If the 10th term of an A.P is 52 and 17th term is 20 more than the 13th term, find A.P..

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

• $\sf a_{10}=52$
• $\sf a_{17}=a_{13}+20$

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

• A.P ?

$\bf{\underline{\underline{Formula:-}}}$

$\bf{\underline{\boxed{\red{a_n=a+(n-1)d}}}}$

where,

• a = First term
• d = common difference

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

$\sf → a_{10}=52$

$\sf → a+9d=52---equ(¡)$

Similarly,

$\sf → a_{17}=a_{13}+20$

$\sf → a+16d=a+12d+20$

$\sf → {\cancel{a}}+16d={\cancel{a}}+12d+20$

$\sf → 16d=12d+20$

$\sf → 16-12d=20$

$\sf → 4d=20$

$\sf→ d=\dfrac{\cancel{20}}{\cancel{4}}$

$\sf {\boxed{\red{d=5}}}$

$\bf{\underline{\underline{Substituting\: value\:of\:"d"\:in\:equ(¡) :-}}}$

$\sf → a+9d=52$

$\sf → a+9×5=52$

$\sf → a = 52-9×5$

$\sf{\boxed{\red{ a_{First\:term} = 7}}}$

$\bf{\underline{\underline{Hence:-}}}$

• The required A.P is
• a , a + d , a + 2d , a + 3d

† A.P = 7 , 12 , 17 , 22 ....