Question

The A.P. in which 4th term is -15 and9th term is -30. Finf the sum of the first 10 numbers.

Answer 1

a4=-15
a9=-30

 a+3d=-15......1
- a+8d=-30......2
  -   -      +
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       -5d=15
           d=-3

Substitute value of d in eqn 1
a+3d=-15
a+3(-3)=-15
a-9=-15
a=-15+9
a=-6.

Sn=n/2+(n-1)d
S10=5+4(-3)
S10=5-12
S10=-7.

Hope it helps u!!!

Answer 2

Hii dear user ....

Answer is here ☺☺............

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Let a be the first term and D be the common difference of the given A.P , than,


T4 = -15

=> a + ( n-1) d = -15
=> a + (4-1) d = -15
=> a + 3d = -15 ........( ¡ )

T 9 = -30

=> a + ( n-1 ) d = -30
=> a + (9-1) d = -30
=> a + 8d = -30. ........( ¡¡ )


On substituting value of d in ( ¡ )

a + 3d = -15
=> a + 3 (-3)=-15
=> a - 9 = -15
=> a = -15 + 9 = -6

$using \: \: formula \: - \\ \\ \frac{s}{n} = (2a + (n - 1)d) \\ \\ = > s10 = 5 + 4( - 3) \\ \\ = > s10 = 5 - 12 \\ \\ = > s10 = - 7$
Here the sum of first number is - 7.

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Hope it's helps you.
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