Question

Which term of an AP: -2, 3, 8,……. is 68 ?

Answer 1

Answer:

first term=-2

d=3-(-2)

d=5

a+(n-1) d

-2+(n-1) 5=68

-2+5n-5=68

-7+5n=68

5n=68-7

5n=61

n=61/5

Answer 2

Answer:

$\huge \color{aqua}\mathbb{\underline {GIVEN } : }$

➡ a = - 2

➡ d = - 5

➡ an = 68

$\huge \mathbf {\boxed {an = a + (n -1) d}}$

Substituting in the formula,

=> 68 = -2 + ( n - 1 ) -5

=> 70 = ( n - 1 ) -5

=> n - 1 = - 14

➡ n = 13

Therefore, the 13 th term of the AP is 68

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