Question

The sum of 4th and 8th terms of an arithmetic progression is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the arithmetic progression.​

Answer 1

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Q) The sum of 4th and 8th terms of an Airihmetic Progression is 24 and the sum of the 6th and 10th term 44 . Find the first three terms of the Airihmetic Progression .

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✿ Concept :

• In these type of Questions , we find the value of first term and Common difference using the given relation in Question .

• In this question , we would make 2 equations using the given relation .

• Then we would solve both the Equations to get the value of (a) and (d) .

• Then using a and d we would find the first 3 terms .

✿ Given :

• 4th term + 8th term = ${a}_4 + {a}_8$ = 24
• 6th term + 10th term = $a_6 + a_{10}$ = 44
• It's an A.P.

✿ To Find :

• The first three terms ($a , a_2 , a_3$)

✿ Solution :

• a = first term
• d = common difference

According to the Question :

$\rm \: 4th \: term + 8th \: term = 24$

$\mapsto \rm \: a _{4} + a _{8} = 24$

$\mapsto \rm \: a + 3d + a + 7d = 24$

$\mapsto \rm \: 2a + 10d = 24$

$\mapsto \boxed{ \rm \: a + 5d = 12}.....(i)$

Again ,

According to the Question :

$\rm6th \: term + 10th \: term = 44$

$\rm \: a _{6} + a _{10} = 44$

$\mapsto \rm \: a + 5d + a + 9d = 44$

$\mapsto \rm \: 2a + 14d = 44$

$\mapsto \boxed{ \rm \: a + 7d = 22}.....(ii)$

Now , we have 2 equation , we can solve it using any method .

Let's use substitution ..

From eq (i) ,

$\rm \: a = 12 - 5d$

Put it in eq(ii)

$\mapsto \rm \: 12 - 5d + 7d = 22$

$\mapsto \rm \: 2d = 22 - 12$

$\mapsto \rm \: \cancel 2d = \cancel{10}$

$\leadsto \boxed{ \sf {\: d = 5}}$

Now ,

$\rm \: a = 12 - 5d$

$\rm \: a = 12 - 5 \times 5$

$\rm \: a = 12 - 25$

$\leadsto \boxed{ \sf \: a = - 13}$

So ,

now we have

• a = -13
• d = 5

So ,

First three terms would be :

• 1st term = a = -13
• 2nd term = $a_2$ = -13+5 = -8
• 3rd term = $a_3$ = -13 + 2×5 = -13 + 10 = -3

So ,

$\frak{first \: 3 \: terms \: are \: \pink{ - 13 \: , - 8 \: and \: - 3}}$

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✯ Know More :

• A.P. = It's a sequence of numbers , in which adjacent terms differ with Common difference .

• $\sf n^{th} \; term = a + (n-1)d$

$\longrightarrow \sf sum \: of \: n \: terms = \frac{n}{2} \{2a + (n - 1)d \}$

Where ,

$\purple{ \ddot{ \smile}} \: \sf\: a = first \: term$

$\green{ \ddot{ \smile}} \: \sf \: d = common \: difference$

$\pink{ \ddot{ \smile}} \: \: \sf \: n = no. \: of \: terms.$

Answer 2

given

t4+t8=24

t6+t10=44

sol.

we know that , in an A.P.

tn= a+(n-1)d

According to given condition

t4+t8=24

a+(4-1)d+a+(8-1)d=24

a+3d+a+7d=24

2a+10d=24.....(1)

According to given condition

t6+t10=44

a+(6-1)d+a+(10-1)d=44

a+5d+a+9d=44

2a+14d=44......(2)

from 1 and 2,

2a+14d=44

- 2a+10d=24

we get d=20/5

d=5

substituting d=5 in equation 1

2a+ 10 × 5= 24

2a= 24 - 50

a= - 13

first three terms

t1 = a= - 13

t2 = a+d= -13 +5= -8

t3 = a+2(d)= -13+ 10= - 3

solution: First three terms of A.P. are -13,-8,-3