Question

determine the ap whose third term is 16 and 7th term exceeds the 5th term by 12​

Answer 1

Answer

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Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

${\boxed{\sf\;{Arithemetic\;Progression}}}$

★Assume the a be first term

★d be the common difference

$\rightarrow{a_{3}=16}$

$\rightarrow{a_{7}-a_{5}}$

= 12

★Here we have :-

a + 2d = 16 ..... (1)

★Also we have :-

(a + 6d) - (a + 4d) = 12

a + 6d - a - 4d = 12

2d = 12

${\boxed{\sf\:{d=\dfrac{13}{2}}}}$

d = 6 ..... (2)

$\fbox{Substitute\;the\;value\;of\;(2) in (1) :-}$

a + 2(6) = 16

a + 12 = 16

a = 16 - 12

a = 4

★Hence AP are :-

(4)

(4 + 6) = 10

(4 + 2 × 6) = 4 + 12 = 16

(4 + 3 × 6) = 4 + 18 = 22

★AP are :-

$\fbox{4 , 10 , 16 , 22}$