Math, asked by reshmashreya, 9 months ago

the sum of first six terms of an AP is 0 and fourth term is 2 find the sum of first 30 terms​

Answers

Answered by VishnuPriya2801
20

Answer:-

Given:

Sum of first six terms of an AP = 0

Fourth term [a(4)] = 2

→ a + 3d = 2

→ a = 2 - 3d -- equation (1).

We know that,

Sum of n terms of an AP = n/2 * [ 2a + (n - 1)d ] .

→ S(6) = 6/2 * [ 2a + 5d ]

→ 0 = 3(2a + 5d)

→ 0 = 6a + 15d

→ 0 = 6(2 - 3d) + 15d

→ 12 - 18d + 15d = 0

→ 12 - 3d = 0

→ 3d = 12

→ d = 12/3

d = 4

Substitute "d" value in equation (1).

→ a = 2 - 3d

→ a = 2 - 3(4)

→ a = 2 - 12

a = - 10

Now,

S(30) = 30/2 * [ 2(-10) + (30 - 1)(4) ]

→ S(30) = 15( - 20 + 116)

→ S(30) = 15(96)

S(30) = 1440

Therefore, the Sum of first 30 terms of the given AP is 1440.

Answered by Anonymous
10

Answer :

Let the first term be ' a ' and common diference of the AP be ' d '

Fourth term a(4) = 2

=> a + ( 4 - 1 )d = 2

=> a + 3d = 2

=> a = 2 - 3d --- Eq( 1 )

Sum of first 6 terms S(6) = 0

=> n / 2 × ( 2a + ( n - 1 )d ) = 0

=> 6 / 2 × ( 2a + ( 6 - 1 )d ) = 0

=> 2a + 5d = 0

From Eq( 1 )

=> 2( 2 - 3d ) + 5d = 0

=> 4 - 6d + 5d = 0

=> 4 = d

=> d = 4

Substituting d = 4 in Eq( 1 )

=> a = 2 - 3( 4 )

=> a = 2 - 12

=> a = - 10

Using Sum of n terms formula

S(n) = n/2 × ( 2a + ( n - 1 )d )

Sum of first 30 terms S(30) = 30/2 × ( 2( - 10 ) + ( 30 - 1 )4 )

= 15 × ( - 20 + 29( 4 ) )

= 15 × ( - 20 + 116 )

= 15 × 96

= 1440

Therefore sum of first 30 terms is 1440.

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