Question

How many terms are there in the A.P.?
$-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},....,\frac{10}{3}$

Answer 1

Ans.

given A.P is

$-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}.....,\frac{10}{3}$

first term of this A.P is $a_1=-1$

second term of this A.P is $a_2=-\frac{5}{6}$

last term of A.P $a_n=\frac{10}{3}$

common difference of an A.P is given by

$(i.e.,d=a_{n+1}-a_n)$

putting n=1 in above equation

$d=a_2-a_1=-\frac{5}{6}-(-1)=\frac{1}{6}$

hence common difference for this A.P is $d=\frac{1}{6}$

the nth term of an A.P is given by.

$a_n=a_1+(n-1)d$

$⟹a_n=-1+(n-1)\frac{1}{6}$

$⟹a_n=-1+\frac{1}{6}n-\frac{1}{6}$

$⟹a_n=-\frac{7}{6}+\frac{1}{6}n...eq\left(1\right)$

for finding number of terms (n) in this A.P, put the last term $a_n=\frac{10}{3}$in eq(1)

$⟹\frac{10}{3}=-\frac{7}{6}+\frac{1}{6}\times n$

$⟹\frac{1}{6}\times n=\frac{10}{3}+\frac{7}{6}$

$⟹\frac{1}{6}\times n=\frac{20+7}{6}$

$⟹n=27$

hence in this A.P number of terms are $n=27$