Question

in an ap consists of 5 terms sum of the first three terms is 21 and the sum of last three terms is 45 find the AP​

Answer 1

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

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Total number of terms in the AP = 5

$\:\:$

• Sum of the first three terms is 21

$\:\:$

• Sum of last three terms is 45

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• The AP

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

$\underline{\bold{\texttt{We know that :}}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$ ----- (1)

$\:\:$

• $\sf S_n$ = Sum of n terms

• n = Number of terms

• a = First term

• d = Common difference

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

Sum of the first three terms is 21

$\:\:$

$\underline{\bold{\texttt{For sum of first three terms :}}}$

$\:\:$

• $\sf S_n$ = 21

• n = 3

• a = a

• d = d

$\:\:$

Putting these values in (1)

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf 21 = \dfrac { 3} { 2 } (2a + (3 - 1)d )$

$\:\:$

➜ $\sf 21 = \dfrac { 3} { 2 } (2a + 2d)$

$\:\:$

➜ $\sf 21 = 3(a + d)$

$\:\:$

➜ $\sf a + d = \dfrac { 21 } { 3 }$

$\:\:$

➠ a + d = 7 ------ (2)

$\:\:$

Given that tere are total 5 term in the AP

$\:\:$

So, Sum of last three term can be found as :

$\:\:$

➠ $\sf S_5 - S_2$

$\:\:$

• $\sf S_5$ = Sum of first 5 terms

• $\sf S_2$ = Sum of first 2 terms

$\:\:$

$\underline{\bold{\texttt{For sum of first 5 terms :}}}$

$\:\:$

• $\sf S_n = S_5$

• n = 5

• a = a

• d = d

$\:\:$

$\underline{\bold{\texttt{Putting these valued in (1) }}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf S_5 = \dfrac { 5 } { 2 } (2a + (5 - 1)d )$

$\:\:$

➜ $\sf S_5 = \dfrac { 5 } { 2 } (2a + 4d)$

$\:\:$

➜ $\sf S_5 = 5(a + 2d)$ ------ (3)

$\:\:$

$\underline{\bold{\texttt{For sum of first 2 terms :}}}$

$\:\:$

• $\sf S_n = S_2$

• n = 2

• a = a

• d = d

$\:\:$

$\underline{\bold{\texttt{Putting these valued in (1) }}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf S_2 = \dfrac { 2 } { 2 } (2a + (2 - 1)d )$

$\:\:$

➜ $\sf S_2 = (2a + d)$ -------- (4)

$\:\:$

We know that,

$\:\:$

➠ $\sf S_5 - S_2 = S_3$

$\:\:$

• $\sf S_5$ = 5(a + 2d)

• $\sf S_2$ = (2a + d)

• $\sf S_3$ = 45

$\:\:$

$\underline{\bold{\texttt{Subtracting (4) from (3) }}}$

$\:\:$

➜ 5(a + 2d) - (2a + d) = 45

$\:\:$

➜ 5a + 10d - 2d - d = 45

$\:\:$

➜ 3a + 9d = 45

$\:\:$

Diving above equation by 3

$\:\:$

➜ a + 3d = 15 ------- (5)

$\:\:$

$\underline{\bold{\texttt{Subtracting (2) from (5) }}}$

$\:\:$

➜ a + 3d - a - d = 15 - 7

$\:\:$

➜ 2d = 8

$\:\:$

➨ d = 4

$\:\:$

• Hence common difference is 4

$\:\:$

$\underline{\bold{\texttt{Putting d = 4 in (2) }}}$

$\:\:$

➜ a + d = 7

$\:\:$

➜ a + 4 = 7

$\:\:$

➜ a = 7 - 4

$\:\:$

➨ a = 3

$\:\:$

• Hence first term is 3

$\:\:$

$\underline{\bold{\texttt{The AP will be :}}}$

$\:\:$

• a + 0d = 3 + 0(4) = 3

• a + 1d = 3 + 1(4) = 7

• a + 2d = 3 + 2(4) = 11

• a + 3d = 3 + 3(4) = 15

• a + 4d = 3 + 4(4) = 19

$\:\:$

• The AP will be 3 , 7 , 11 , 15 , 19

Answer 2

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

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Total number of terms in the AP = 5

$\:\:$

• Sum of the first three terms is 21

$\:\:$

• Sum of last three terms is 45

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• The AP

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

$\underline{\bold{\texttt{We know that :}}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$ ----- (1)

$\:\:$

$\sf S_n$ = Sum of n terms

n = Number of terms

a = First term

d = Common difference

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

Sum of the first three terms is 21

$\:\:$

$\underline{\bold{\texttt{For sum of first three terms :}}}$

$\:\:$

$\sf S_n$ = 21

n = 3

a = a

d = d

$\:\:$

Putting these values in (1)

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf 21 = \dfrac { 3} { 2 } (2a + (3 - 1)d )$

$\:\:$

➜ $\sf 21 = \dfrac { 3} { 2 } (2a + 2d)$

$\:\:$

➜ $\sf 21 = 3(a + d)$

$\:\:$

➜ $\sf a + d = \dfrac { 21 } { 3 }$

$\:\:$

➠ a + d = 7 ------ (2)

$\:\:$

• Given that tere are total 5 term in the AP

$\:\:$

So, Sum of last three term can be found as :

$\:\:$

➠ $\sf S_5 - S_2$

$\:\:$

$\sf S_5$ = Sum of first 5 terms

$\sf S_2$ = Sum of first 2 terms

$\:\:$

$\underline{\bold{\texttt{For sum of first 5 terms :}}}$

$\:\:$

$\sf S_n = S_5$

n = 5

a = a

d = d

$\:\:$

$\underline{\bold{\texttt{Putting these valued in (1) }}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf S_5 = \dfrac { 5 } { 2 } (2a + (5 - 1)d )$

$\:\:$

➜ $\sf S_5 = \dfrac { 5 } { 2 } (2a + 4d)$

$\:\:$

➜ $\sf S_5 = 5(a + 2d)$ ------ (3)

$\:\:$

$\underline{\bold{\texttt{For sum of first 2 terms :}}}$

$\:\:$

$\sf S_n = S_2$

n = 2

a = a

d = d

$\:\:$

$\underline{\bold{\texttt{Putting these valued in (1) }}}$

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d )$

$\:\:$

➜ $\sf S_2 = \dfrac { 2 } { 2 } (2a + (2 - 1)d )$

$\:\:$

➜ $\sf S_2 = (2a + d)$ -------- (4)

$\:\:$

We know that,

$\:\:$

➠ $\sf S_5 - S_2 = S_3$

$\:\:$

$\sf S_5$ = 5(a + 2d)

$\sf S_2$ = (2a + d)

$\sf S_3$ = 45

$\:\:$

$\underline{\bold{\texttt{Subtracting (4) from (3) }}}$

$\:\:$

➜ 5(a + 2d) - (2a + d) = 45

$\:\:$

➜ 5a + 10d - 2d - d = 45

$\:\:$

➜ 3a + 9d = 45

$\:\:$

• Diving above equation by 3

$\:\:$

➜ a + 3d = 15 ------- (5)

$\:\:$

$\underline{\bold{\texttt{Subtracting (2) from (5) }}}$

$\:\:$

➜ a + 3d - a - d = 15 - 7

$\:\:$

➜ 2d = 8

$\:\:$

➨ d = 4

$\:\:$

• Hence common difference is 4

$\:\:$

$\underline{\bold{\texttt{Putting d = 4 in (2) }}}$

$\:\:$

➜ a + d = 7

$\:\:$

➜ a + 4 = 7

$\:\:$

➜ a = 7 - 4

$\:\:$

➨ a = 3

$\:\:$

Hence first term is 3

$\:\:$

$\underline{\bold{\texttt{The AP will be :}}}$

$\:\:$

a + 0d = 3 + 0(4) = 3

a + 1d = 3 + 1(4) = 7

a + 2d = 3 + 2(4) = 11

a + 3d = 3 + 3(4) = 15

a + 4d = 3 + 4(4) = 19

$\:\:$

• The AP will be 3 , 7 , 11 , 15 , 19