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

Given a10=52

an=a+(n−1)d52=a+9d(1)

also, a17=20+a13a+16d=20+a+12d4d=20d=5(2)

From (1) and (2)

52=a+9×5a=7

The AP is: 7,12,17,22,...

Hope it helps you....

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

•$\sf a_{10}=52$

•$\sf a_{17}=a_{13}+20$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• A.P

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{ Here, we know the formula :}}$

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

Where,

• a = First term

• d = common difference

$\bf\underline{\green{\:\:\:\:\:\:\:\:A.T.Q:-\:\:\:\:\:\:\:}}$

$\sf → a_{10}=52$

$\sf → a+9d=52---eq(¡)$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Again,

$\sf → a_{17}=a_{13}+20$

$\sf → a+16d=a+12d+20$

$\sf → 16d=12d+20$

$\sf → 4d=20$

$\sf → d= 5$

$\underline{\:\textsf{Now, put the value of "d" :}}$

$\sf → a+9d=52$

$\sf → a+9×5=52$

$\sf → a = 52-9×5$

$\sf → a= 7$

$\bf{\underline{\underline{Now-}}}$

The required A.P is _

a , a + d , a + 2d , a + 3d.........

Where,

• a = 7

• a + d = 5 + 7 = 12

• a + 2d = 17

• a + 3d = 22

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

• A.P = 7 , 12 , 17 , 22 ....

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