Question

if the 10th term of an AP is 52 and 16th team is 18more than 13th term find the AP
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Answer 1

Given :

• $\sf {10}^{th}\:term\:=\:52$
• $\sf {16}^{th}\:term\:is\:18\:more\:than\:{13}^{th}\:term$

To Find :

• A.P.

Solution :

$\large\bold{\boxed{\sf{\red{a_n\:=\:a\:+\:(n\:-\:1)d}}}}$

• According to the question :

• It is given that $\sf {10}^{th}$ is 52

$\leadsto \sf 52 \:=\: a\:+\:(10\:-\:1)d$

$\leadsto \sf 52\:=\: a\:+\:9d$

• It is also given that $\sf {16}^{th}$ term is 18 more than $\sf {13}^{th}$ term

$\leadsto \sf {a}_{16} \:=\: {a}_{13} \:+\:18$

$\leadsto \sf a\:+\:15d\:=\: a\:+\:12d\:+\:18$

$\leadsto \sf 15d\:-\:12d\:=\:18$

$\leadsto \sf 3d\:=\:18$

$\leadsto \sf d\:=\: \dfrac{18}{3}$

$\leadsto \sf d\:=\: 6$

• Common difference is 6 , now , we have to find first term :

$\leadsto \sf 52\:=\:a\:+\:9(6)$

$\leadsto \sf -a\:=\:54\:-\:52$

$\leadsto \sf a\:=\:-2$

• First term of the A.P. is -2

• A.P. :

• First term = -2
• Second term = -2 + (2 - 1)6 = 4
• Third term = -2 + (3 - 1)6 = 10
• Fourth term = -2 + (4 - 1)6 = 16

• A.P. = -2 , 4 , 10 , 16 ...

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