Question

Which term of A.P. 5/6, 1, 1 1/6, 3 1/3 is 3?

Answer 1

Answer:

$elloh$

A=5/6

D=t2-t1

=1-5/6

=1/6

Let 3 be the tn term of this ap

Tn=a+(n-1)d

3=5/6+(n-1)1/6

3-5/6=n-1)1/6

13/6 ×6=n-1

13=n-1

14=n

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{14\:term\:of\:A.P\:is\:3}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Fiven : }} \\ \tt: \implies A.P = \frac{5}{6} ,1, \frac{7}{6} , \frac{10}{3} \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Which \: term \: is \: 3 = ?$

• According to given question :

$\tt \circ \: First \: term = \frac{5}{6} \\ \\ \tt : \implies Common \: difference = a_{2} - a_{1} \\ \\ \tt : \implies Common \: difference =1 - \frac{5}{6} \\ \\ \tt \circ \: Common \: difference = \frac{1}{6} \\ \\ \tt \circ \: a_{n} = 3 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\ \tt: \implies 3 = \frac{5}{6} + (n - 1) \times \frac{1}{6} \\ \\ \tt: \implies 3 - \frac{5}{6} = \frac{1}{6} n - \frac{1}{6} \\ \\ \tt: \implies \frac{18 - 5}{6} = \frac{n - 1}{6} \\ \\ \tt: \implies 18 - 5 = n - 1 \\ \\ \tt: \implies 13 + 1 = n \\ \\ \green{\tt: \implies n = 14 } \\ \\ \green{\tt \therefore 14 \: th \: term \: of \: this \: ap \: is \: 3}$