Question

The 15th term from the end in the AP 13 ,16 ,19 ,.......,160 is​

Answer 1

$Given \:A.P: 13,16,19, \ldots ,160$

$Common \: difference \: of \: A.P$

$= a_{2} - a_{1}$

$= 16 - 13$

$= 3$

/* Now, Rearranging the given A.P in reverse order , it becomes */

$New \: A.P: 160, 157, 154,\ldots ,19,16,13$

$First\:term (a) = 160$

$Common \: difference (d) = a_{2} - a_{1}$

$= 157 - 160$

$= -3$

$\boxed{\pink{ n^{th} \:term (a_{n}) = a+(n-1)d }}$

$Here ,a = 160, d = -3 \:and \: n = 15$

$\red{ 15^{th} \:term \:in \:A.P }$

$\implies a_{15} = a + 14d$

$= 160 + 14\times (-3)$

$= 160 - 42$

$= 118$

Therefore.,

$\red{ 15^{th}\: term \:from \:the \:end \:of }$

$\red{ given \:A.P}\green{= 118}$

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Answer 2

Answer:

here is your answer

Step-by-step explanation:

Answer:- 118