Question

the sum of 4th and 8th terms of an AP is 37 and the sum of 6th and 12th term is 46 .Find the first term of an AP?

Answer 1

Answer:

Given that Sum of 4th term and 8th term of an AP is 37. Also Sum of 6th term and 12th term is 46.

We must find the first term of the given AP.

4th term of an AP cana be written : a + 3d.

6th term = a + 5d, 8th term = a + 7d, 12th term = a + 11d.

Substituting the values we get,

=> a + 3d + a + 7d = 37

=> 2a + 10d = 37   ...( 1 )

Similarly,

=> a + 5d + a + 11d = 46

=> 2a + 16d = 46

=> a + 8d = 23   ...( 2 )

Solving ( 1 ) and ( 2 ), we get,

a + 8d = 23

=> a = 23 - 8d   ...( 3 )

Substituting the value of 'a' from ( 3 ) in ( 1 ) we get,

=> 2a + 10d = 37

=> 2 ( 23 - 8d ) + 10d = 37

=> 46 - 16d + 10d = 37

=> 46 - 6d = 37

=> 46 - 37 = 6d

=> 9 = 6d

=> d = 9 / 6

=> d = 3 / 2

Now substituting value of d in ( 3 ) , we get,

=> a = 23 - 8d

=> a = 23 - 8 ( 3 / 2 )

=> a = 23 - ( 24 / 2 )

=> a = 23 - 12

=> a = 11

Hence the first term of the AP is ( 11 ).

Answer 2

✨here is ur answer ↓✨

a4+a8=37✔
(a+3d)+(a+7d)=37✔
2a+10d=37✔
2(a+5d)=37. ✔ (1). ✔✔✔✔


a6+a12=46✔
(a+5d)+(a+11d)=46✔
2a+16d=46✔
2(a+8d)=46✔
a+8d=46/2✔
a+8d=23✔
a=23-8d.✔. (2)✔✔✔✔
replace value of "a" in equation (1)
2{(23-8d)+5d} =37
2(23-3d)=37
46-6d=37
-6d=37-46
-6d=-9
d=-9/-6
d=3/2✔✔
replace d in equation (2)
a=23-8(3/2)
a=23-(4×3)
a=23-12
a=11✔✔

hence 1st term of AP is 11✔✔

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