Question

The sum of first 11terms in an arithmetic sequence is 275.the sum of first 19 terms is 779.what is the 6th term

Answer 1

Answer:

25

Step-by-step explanation:

Let the first term be a and com. diff. be d. Using S = (n/2)[2a + (n - 1)d]

=> sum of 11 terms = 275

=> (11/2) (2a + 10d) = 275

=> 11(a + 5d) = 275

=> a + 5d = 25 ...(1)

Sum of 19 terms = 779

=> (19/2)(2a + 18d) = 779

=> a + 9d = 41 ...(2)

Subtract (1) from (2) : we get

=> 4d = 16

=> d = 4

Thus, eqⁿ(1) is a + 5(4) = 25 => a = 5

Hence, using nth term = a + (n - 1)d

6th term = 5 + (6 - 1)4 = 5 + (5)4

6th term = 25

Answer 2

Given that , The sum of first 11 terms in an arithmetic sequence is 275 & the sum of first 19 terms is 779 .

Exigency To Find : The 6 th term of an A.P ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

¤ To Calculate the Sum total of an A.P is Given by ━━

$\qquad \star\:\underline{\boxed {\pmb{\sf{ S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup}}}}$

⠀⠀⠀⠀⠀Here n is the n th Term of an A.P a is the first term of an A.P & d is common Difference of an A.P.

⠀⠀CAEE I : The sum of first 11 terms in an arithmetic sequence is 275 .

$\qquad \dashrightarrow \sf S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: \dfrac{11}{2} \bigg\lgroup \sf{ 2a + ( 11 - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: \dfrac{11}{2} \bigg\lgroup \sf{ 2a + 10 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: 11 \bigg\lgroup \sf{ a + 5 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup$

⠀⠀CAEE II : The sum of first 19 terms in an arithmetic sequence is 779 .

$\qquad \dashrightarrow \sf S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: \dfrac{19}{2} \bigg\lgroup \sf{ 2a + ( 19 - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: \dfrac{19}{2} \bigg\lgroup \sf{ 2a + 18 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: 19 \bigg\lgroup \sf{ a + 9 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf a + 9d\: =\: 41 \:\qquad \bigg\lgroup \sf{ eq^n \:ii }\bigg\rgroup$

We get ,

$\qquad \leadsto \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup$

$\qquad \leadsto \sf a + 9d\: =\: 41 \:\qquad \bigg\lgroup \sf{ eq^n \:ii }\bigg\rgroup$

$\dag \underline {\frak {By \: Subtracting\: eq^n\:i \: from \: eq^n \:ii\::}}$

$\qquad \dashrightarrow \sf a + 9d\:- ( a + 5d ) =\: 41 \:- 25$

$\qquad \dashrightarrow \sf a + 9d\:- a - 5d =\: 16$

$\qquad \dashrightarrow \sf 4d =\: 16$

$\qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ d \:( \: or \:Common \:Difference\:)\:=\: 4 \:}}}}$

$\dag \underline {\frak{ By \: Substituting \: \:value \: of \: d \: eq^n\: i\::}}$

$\qquad \leadsto \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup$

$\qquad \dashrightarrow \sf a + 5d\: =\: 25 \:$

$\qquad \dashrightarrow \sf a + 5(4)\: =\: 25 \:$

$\qquad \dashrightarrow \sf a \: =\: 25 -20 \:$

$\qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ a \:( \: or \:First \:Term\:)\:=\: 5 \:}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀━ Finding 6 th term of an A.P :

¤ To Calculate an n th terms of an A.P is Given by ━

$\qquad \star\:\underline{\boxed {\pmb{\sf{ a_n \: =\: a + ( n - 1 ) d }}}}$

⠀⠀⠀⠀⠀Here n is the n th Term of an A.P a is the first term of an A.P & d is common Difference of an A.P.

$\qquad \dashrightarrow \sf a_n \: =\: a + ( n - 1 ) d \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + ( 6 - 1 ) 4 \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + ( 5 ) 4 \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + 20 \\\\ \qquad \dashrightarrow \sf a_n \: =\: 25 \\\\ \qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ a_6 \:( \: or \:Sixth \:Term\:)\:=\: 25 \:}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:6^{th}\:term\:of\:an\:A.P \:is\:\pmb{\bf 25\:}\:.}$