Question

if a1, a2, a3........ ... an forms an A.P and an = 3+4n then find the fifteen term *



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Answer 1

Answer:

an = 3+4n

a15= 3+ 4(15)

a15 = 63

Therefore 15th term of the AP is 63..

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Answer 2

Answer: The A.P. is 7,11,15...63... and its 15th term is 63.

Given:

$\sf a_n=3+4n$

To find:

$\textsf{The A.P.}$

Formula used:

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

$\sf where,$

$\sf a_n=The\:n^{th}\:term.$

$\sf a=The\:first\:term.$

$\sf n=Total\:number\:of\:terms.$

$\sf d=The\:common\:difference.$

Explanation:

$\textsf{According to the question,}$

$\sf a_n=3+4n$

$\textsf{Putting n=1, we get:}$

$\Rightarrow \sf a_1=3+4(1)$

$\Rightarrow \sf a_1=7$

$\textsf{Putting n=2, we get:}$

$\Rightarrow \sf a_2=3+4(2)$

$\Rightarrow \sf a_2=11$

$\textsf{Putting n=3, we get:}$

$\Rightarrow \sf a_3=3+4(3)$

$\Rightarrow \sf a_3=15$

.

.

.

$\textsf{Putting n=15, we get:}$

$\Rightarrow \sf a_3=3+4(15)$

$\Rightarrow \sf a_3=63$

$\therefore \textsf{The A.P. is 7,11,15...63...}$

Hence, the A.P. is 7,11,15...63... and its 15th term is 63.

More to know:

Arithmetic Progression (AP): A sequence of numbers in which the difference of two consecutive numbers is a constant.