Question

The first, second and the last terms of an A.P. are 101, 111 and 2001. Find the number of terms.

Answer 1

hey mate here is ur answer

a=101

d=111-101=10

tn=a+(n-1)d

2001=101+(n-1)10

2001=101+10n-10

2001=91+10n

10n=1910

n=191

hence number of terms =191.

hope it helps.

☆☆☆

Answer 2

${ \bold{ \boxed{ \boxed{ \green{ \huge{ \star \: ANSWER \: \star}}}}}}$

${ \bold{ \huge \: \underline{Given} : - }} \\ \\ { \bold{ \red{ \implies \: first \: \: term \: \: of \:A.P. = 101}}} \\ \\ { \bold{ \red{ \implies \:second \: \: term \: \: of \: \: A.P. = 111}}} \\ \\ { \bold{ \red{ \implies \: last \: \: term \: \: of \: \:A.P. = 2001}}} \\ \\ \\ \\ { \bold{ \underline{ \huge{ To \: \: find }} : - }} \\ \\ { \bold{ \red{ \implies \: total \: \: number \: \: of \: \: terms \: \: in \: \: A.P. \: }}} \\ \\ \\ { \bold{ \underline{ \huge{used \: \: formula} } : - }} \\ \\ { \bold{ \red{ \implies{ \boxed{ T_{n} = a + (n - 1)d }} }}}\\ \\ \: \: \: \: { \bold{ \blue{ \: \: \: \: \: \: . \: where \: \: a = first \: \: term }}}\\ \\ { \bold{ \blue{ \: \: \: \: \: \: \: \: \: \: . \: \: d = common \: \: difference\:}}}\\ \\ { \bold{ \blue{ \: \: \: \: \: \: \: \: \: \: \: . \: \:n = total \: \: nuber \: \: of \: \: terms}}}\\ \\{ \bold{ \blue{ \: \: \: \: \: \: \: \: \: \: \: . \:T_{n} = last \: \: term}}}\\ \\ \\ { \bold{ \boxed{ \boxed{ \huge{ \green{ \bigstar \: solution \: \bigstar}}}}}}$

${ \bold{ \red{ \implies \:T _{n} = a + (n - 1)d }}} \\ \\ \\ { \bold{ \red{ \implies \: 2001 = 101 + (n - 1)d}}} \\ \\ { \bold{ \blue{ \: \: . \: now \: \: common \: \: difference = 111 - 101 = 10}}} \\ \\ \\ { \bold{ \red{ \implies \:2001 = 101 + (n - 1)10 }}} \\ \\ \\ { \bold{ \red{ \implies \: 1900 = (n - 1)10}}} \\ \\ { \bold{ \red{ \implies \: (n - 1) = 190}}} \\ \\ \\ { \bold{ \red{ \implies \:{ \boxed{n = 191}} }}}$