Question

4th term of an A.P. is 16 and the 9th term is 31, find the first 5 terms

Answer 1

$\large\bf\underline{Given:-}$

• 4th term = 16
• 9th term = 31

$\large\bf\underline {To \: find:-}$

• first five terms of AP.

$\huge\bf\underline{Solution:-}$

$\large \bf \: T_n = a+(n-1)d \\ \rm \: T_{4} = a+3d \\ \rm \: T_9 = a + 8d$

$\large \bf \: T_n = a+(n-1)d \\ \\ \rm \: T_{4} = a+3d \\ \\ \rm \dashrightarrow \rm \: a + 3d = 16 ........(i)\\ \\ \dashrightarrow \rm \:T_{9} = a+8d \\ \\ \: \dashrightarrow \rm \: a + 8d = 31.......(ii)$

From (i) and (ii)

$\boxed{ \begin{minipage}{4cm} a + 3d = 16 \\ a + 8d = 31 \\ - \: \: \: - \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: } \\ \: . \: - 5d \: \: = - 15 \\ . \: \: d \: \: = \bf 3 \end{minipage}}$

Putting value of d = 3 in equation (i)

$\rm \dashrightarrow \: a + 3d = 16 \\ \\ \rm \dashrightarrow \: a + 3 \times 3 = 16 \\ \\ \rm \dashrightarrow \: a + 9 = 16 \\ \\ \rm \dashrightarrow \: a = 16 - 9 \\ \\ \rm \dashrightarrow \: a = 7$

• common difference (d) = 3
• First term (a) = 7

$\boxed{ \begin{minipage}{4cm}first \: term \: = a \\ second \: term = a + d \\ third \: term = a + 2d \\ fourth \: term \: a + 3d \\ fifth \: term \: = a + 4d\end{minipage}}$

So,

First term = 7

second term = 7+3 = 10

third term = 7+ 2×3 = 13

fourth term = 7+3×3 = 16

fifth term = 7+4×3 = 19

≫ the first 5 terms are :- 7 , 10, 13 ,16 ,19.

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ first \ five \ terms \ of \ the \ AP}$

$\sf{are \ 7, \ 10, \ 13, \ 16 \ and \ 19.}$

$\sf\orange{Given:}$

$\sf{\implies{t_{4}=16}}$

$\sf{\implies{t_{9}=31}}$

$\sf\pink{To \ find:}$

$\sf{First \ five \ terms \ of \ an \ AP.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{tn=a+(n-1)d}}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{t_{4}=a+(4-1)d}$

$\sf{\therefore{16=a+3d}}$

$\sf{\therefore{a+3d=16...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{t_{9}=a+(9-1)d}$

$\sf{\therefore{31=a+8d}}$

$\sf{\therefore{a+8d=31...(2)}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{a+8d=31}$

$\sf{-}$

$\sf{a+3d=16}$

___________________

$\sf{5d=15}$

$\sf{\therefore{d=\frac{15}{5}}}$

$\boxed{\sf{\therefore{d=3}}}$

$\sf{Substitute \ d=3 \ in \ equation (1)}$

$\sf{a+3(3)=16}$

$\sf{\therefore{a+9=16}}$

$\sf{\therefore{a=16-9}}$

$\boxed{\sf{\therefore{a=7}}}$

$\sf{The \ required \ terms \ are:}$

$\sf{\implies{t_{1}=a=7,}}$

$\sf{\implies{t_{2}=a+d=7+3=10,}}$

$\sf{\implies{t_{3}=a+2d=7+2(3)=13,}}$

$\sf{\implies{t_{4}=a+3d=7+3(3)=16,}}$

$\sf{\implies{t_{5}=a+4d=7+4(3)=19}}$

$\sf\purple{\tt{\therefore{The \ first \ five \ terms \ of \ the \ AP}}}$

$\sf\purple{\tt{are \ 7, \ 10, \ 13, \ 16 \ and \ 19.}}$