Question

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

Answer 1

Formula Applied :

$\sf \bigstar\ \; a_n=a+(n-1)d$

$\sf First\ term,a_1=a\\\\\sf Second\ term,a_2=a+d\\\\\sf Third\ term,a_3=a+2d$

Explanation :

A/c , " sum of the 4th and 8th terms of an arithmetic progression is 24 "

$\to \sf a_4+a_8=24$

$\to \sf \{a+3d\}+\{a+7d\}=24\\\\\to \sf 2a+10d=24\\\\\to \sf a+5d=12...(1)$

A/c , " sum of 6th and 10th term is 44 "

$\to \sf a_6+a_{10}=44$

$\to \sf \{a+5d\}+\{a+9d\}=44\\\\\to \sf 2a+14d=44\\\\\to \sf a+7d=22...(2)$

Solve (2) - (1) ,

$\sf \to 2d=10\\\\\to \sf d=5\ \; \bigstar$

Sub. d value in (1) ,

$\to \sf a+5(5)=12\\\\\to \sf a=-13\ \; \bigstar$

So ,

First term , a = - 13  $\orange{\bigstar}$

Second term , a + d = - 13 + 5 = - 8  $\bigstar$  

Third term , a + 2d = - 13 + 10 = - 3  $\green{\bigstar}$

Answer 2

Given :

• Sum of 4th term and 8th term of AP = 24

• sum of 6th and 10th term = 44

To find :

• first three terms of the AP

Formula used :

• Tn = a+(n-1)d

where :-

• a = first term
• n = number of terms
• d = common difference
• Tn = nth term

Solution :

4th term ( T4) = a+(4 -1 )d

⟹ T4= a+3d

8th term ( T8) = a+(8-1 )d

⟹ T8= a+7d

Now , sum of T4 and T 8 :-

⟹ T4 + T8 = 24

⟹ a+3d + a+7d = 24

⟹ 2a+10d = 24

Now take out 2 as common:-

⟹ 2(a+5d ) = 24

⟹ (a+5d ) = 24/2

⟹ a+5d = 12

⟹ a = 12-5d..... [1]

6th term ( T6) = a+(6 -1 )d

⟹ T6= a+5d

10th term ( T10) = a+(10-1 )d

⟹ T10= a+9d

Now , sum of T6 and T10 :-

⟹ T6+ T10= 44

⟹ a+5d+ a+9d= 44

⟹ 2a+14d = 44

Now take out 2 as common:-

⟹ 2(a+7d ) = 44

⟹ (a+7d ) = 44/2

⟹ a+7d = 22

⟹ a = 22-7d..... [2]

From [1] and [2] we get :-

⟹ 12-5d= 22-7d

⟹ 7d -5d = 10

⟹ 2d = 10

⟹ d = 10 / 2

⟹ d = 5

Now to find a , put d = 5 in a = 12-5d :-

⟹ a = 12-5(5)

⟹ a = 12-25

⟹ a = -13

Now we have to find first three terms of AP i.e :-

⟹ a , a+d , a+2d

⟹ -13 , -13+5 , -13+2(5)

⟹ -13 , -8 , -3

Answer :-

First three terms of A.P = -13 , -8 , -3 ....

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Learn more:-

1. Tn = a+(n-1)d

2. Sn = n/2[2a+(n-1)d]

3. Sn = n/2(a+l)

WHERE :-

• a = first term
• n = number of terms
• d = common difference
• Tn = nth term
• l = last term

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