Question

which term of the AP : 3, 8, 13, 18, ...., is 78?​

Answer 1

Step-by-step explanation:

Thee given AP is 3, 8, 13, 18,......

Here, 1st term (a) = 3

And, common difference (d) = 8-3 = 13-8 = 5

Let 78 is the nth term of the AP.

So,

$a + (n - 1)d = 78$

=> 3 + (n-1)5 = 78 [ a=3 and d=5 ]

=> (n-1)5 = 78 - 3

=> (n-1)5 = 75

=> (n‐1) = 75÷5

=> n‐1 = 15

=> n = 15+1

=> n = 16

Hence 78 is the 16th term of the A.P..

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Answer 2

$\sf \Large Solution!!$

$\sf \large Given:$

$\sf \to First\:term(a)=3$

$\sf \to Common\:difference(d)=8-3=5$

$\sf \to a_{n}=78$

$\sf \large Formula:$

$\sf \to a_{n}=a+(n-1)d$

$\sf \large So,$

$\sf \to 78=3+(n-1)\times 5$

$\sf \to 78-3=(n-1)\times 5$

$\sf \to 75=(n-1)\times 5$

$\sf \to \dfrac{\cancel{75}\:^{15}}{\cancel 5}=n-1$

$\sf \to 15=n-1$

$\sf \to n=16$

$\sf \bigstar So,\:16^{th}\:term\:of\:an\:A.P.\:is\:78.$

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