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