In the AP find the missing term. _,13,_,3
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Answered by
18
let 2nd term that is A2 = a+(2-1)d = 13
= a+ d = 13 _eq.1
4th term that is A4 = a+(4-1)d = 3
= a+3d = 3_eq.2
now,
eliminating both 1 and 2 eq.
a+d=13
a+3d=3
- - = -
-2d = 10
d=-5
now put the value of d in eq.1
a+(-5)=13
a-5=13
a=18
now third term = A3 = a+2d
= 18+2 (-5)
= 18-10 = 8
so therefore , the A.P. is 18 , 13 , 8 , 3
hope it helps .
hope u will mark my answer as brainliest .
= a+ d = 13 _eq.1
4th term that is A4 = a+(4-1)d = 3
= a+3d = 3_eq.2
now,
eliminating both 1 and 2 eq.
a+d=13
a+3d=3
- - = -
-2d = 10
d=-5
now put the value of d in eq.1
a+(-5)=13
a-5=13
a=18
now third term = A3 = a+2d
= 18+2 (-5)
= 18-10 = 8
so therefore , the A.P. is 18 , 13 , 8 , 3
hope it helps .
hope u will mark my answer as brainliest .
Welcome
and thankyou for marking my answer as brainliest
Answered by
5
Let a1 = first term
d = common difference
a2 = 13
=>a + d = 13 ———> (i)
a4 = 3
=> a+ 3d = 3 ———> (ii)
(i) – (ii)
a + d = 13
a + 3d = 3
(-) (-) (-)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
- 2d = 10
=> d = 10/-2
=> d = - 5
Putting d = - 5 in equ'n (i), we get
a + (- 5) = 13
=> a - 5 = 13
=> a = 13 + 5
=> a = 18
Therefore, a1 = 18
and a3 = a + 2d
= 18 + 2× (- 5)
= 18 - 10
= 8
Hence, the A.P is 18, 13, 8, 3
Hope it'll help you....:)
d = common difference
a2 = 13
=>a + d = 13 ———> (i)
a4 = 3
=> a+ 3d = 3 ———> (ii)
(i) – (ii)
a + d = 13
a + 3d = 3
(-) (-) (-)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
- 2d = 10
=> d = 10/-2
=> d = - 5
Putting d = - 5 in equ'n (i), we get
a + (- 5) = 13
=> a - 5 = 13
=> a = 13 + 5
=> a = 18
Therefore, a1 = 18
and a3 = a + 2d
= 18 + 2× (- 5)
= 18 - 10
= 8
Hence, the A.P is 18, 13, 8, 3
Hope it'll help you....:)
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