Question

write an A . p . whose first is 9 and common difference is 4​

Answer 1

$\maltese\:\underline{\bf AnsWer :}\:\maltese$

In the question we are given that, first term of an AP is 9 and common difference (d) is 4. We need to find the required A.P .

$\bullet\:\sf First \: Term \: (a_1) = 9$

$\bullet\:\sf Common \: Difference \: (d) = 4$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\qquad \quad \dag\: \underline{\textbf{Second Term : }}$

$:\implies \sf Second \: Term = First \: Term + Common \: Difference$

$:\implies \sf a_2 = a_1 + d$

$:\implies \sf a_2 = 9 + 4$

$:\implies \underline{ \overline{ \sf a_2 = 13}}$

$\qquad \quad \dag\: \underline{\textbf{Third Term : }}$

$:\implies \sf a_3 = a_1 + 2d$

$:\implies \sf a_3 = 9+ 2 \times 4$

$:\implies \sf a_3 = 9+ 8$

$:\implies \underline{ \overline{ \sf a_3 = 17}}$

$\qquad \quad \dag\: \underline{\textbf{Fourth Term : }}$

$:\implies \sf a_4 = a_1 + 3d$

$:\implies \sf a_4 = 9+ 3 \times4$

$:\implies \sf a_4 = 9+ 12$

$:\implies \underline{ \overline{ \sf a_4 = 21}}$

$\therefore\:\underline{\textsf{The required A.P is \textbf{9, 13, 17, 21....}}}.\:\dag$

Answer 2

${ \underline{ \underline{ \huge{ \blue{ \tt{SOLUTION}}}}}}$

${ \dag{ \underline{ \underline{ \tt{ \orange{ \: \: given}}}}}} \\ { \dashrightarrow{ \sf{ \purple{ \: \: common \: difference(d) = 4}}}} \\ { \dashrightarrow{ \sf{ \purple{ \: \: first \: term(a) = 9}}}}$

${ \dag{ \underline{ \underline{ \orange{ \tt{ \: \: to \: find}}}}}} \\ { \sf{write \: AP \: series \: from \: the \: given \: data}}$

${ \dag{ \underline{ \underline{ \tt{ \orange{ \: \: \: formula \: used}}}}}} \\ \\ { \boxed{ \boxed{ \tt{ \green{Tn \: = a + (n - 1)d}}}}} \\ { \sf{where}} \\ { \sf{Tn \: = \: { \green{n \: th \: term}}}} \\ { \sf{ a \: = { \green{ \: first \: term}}}} \\ { \sf{n \: = \: { \green{number \: of \: terms}}}} \\ { \sf{d \: = { \green{ \: common \: difference}}}}$

${ : \implies{ \underline{ \red{ \sf{ \: first \: term \: = \: 9}}}}}$

${ : { \implies{ \underline{ \sf{ \red{ \: second \: term \: ( \: T2 \: )}}}}}} \\ { \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = 2}} \\ \: \: \: \: \: \: { \rightarrow{ \sf{ \blue{ \: \: \: T2 = 9 + (2 - 1)4}}}} \\ \: \: \: \: \: \: { \sf{ \rightarrow{ \blue{ \: \: \: T2 = 9 + 4}}}} \\ \: \: \: \: \: \: { \rightarrow \: \: \: { \underline{ \green { \boxed{ \blue{ \sf{ \: T2 = 13 }}}}}}}$

${ : { \implies{ \underline{ \sf{ \red{third \: term \: ( \:T3 \: )}}}}}} \\ { \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = 3}} \\ \: \: \: \: \: \: { \rightarrow{ \blue{ \sf{ \: \: T3 = 9 + (3 - 1)4}}}} \\ \: \: \: \: \: \: { \rightarrow{ \blue{ \sf{ \: \: T3 = 9 + (2 \times 4)}}}} \\ \: \: \: \: \: \: { \rightarrow{ \blue{ \sf{ \: \: T3 = 9 + 8}}}} \\ \: \: \: \: \: \: { \rightarrow \: \: { \underline { \green{ \boxed{ \sf{ \blue{ T3 = 17}}}}}}}$

${ : { \implies{ \underline{ \red{ \sf{fourth \: term \: ( \: T4 \: )}}}}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf{n = 4}} \\ \: \: \: \: \: \: { \rightarrow{ \sf{ \blue{ \: \: T4 = 9 + (4 - 1)4}}}} \\ \: \: \: \: \: \: { \rightarrow{ \sf{ \blue{ \: \: T4 = 9 + (3 \times 4)}}}} \\ \: \: \: \: \: \: { \rightarrow{ \sf{ \blue{ \: \: T4 = 9 + 12}}}} \\ \: \: \: \: \: \: { \rightarrow \: \: { \underline{ \green{ \boxed{ \sf{ \blue{T4 = 21}}}}}}}$

${ : { \implies{ \underline{ \sf{ \red{fifth \: term \: ( \: T5 \: )}}}}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf{n = 5}}$

${ \rightarrow{ \blue{ \sf{ \: \: T5 = 9 + (5 - 1)4}}}} \\ { \rightarrow{ \blue{ \sf{ \: \: T5 = 9 + (4 \times 4)}}}} \\ { \rightarrow{ \blue{ \sf{ \: \:T5 = 9 + 16}}}} \\ { \rightarrow \: \: \: { \underline{ \green { \boxed{ \blue{ \sf{T5 = 25}}}}}}}</p><p>$

The required A.P. is 9, 13, 17, 21, 25,.......