the sum of first term of an AP is given by 4n^2+2n. find nth term of AP
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2
Sn = 4n^2 +2n
=> n/2 [ 2a +(n-1)d] = n(4n+2)
=> 2a +(n-1)d = 8n + 4
=> 2a + (n-1) d = 12 - 8 + 8n
=> 2a + (n-1)d = 12 + 8( n-1)
On Comparing both sides, we get
a = 6
d = 8
Tn = a + (n-1)d
= 6 + (n-1) 8
= 6 + 8n - 8
= 8n - 2
=> n/2 [ 2a +(n-1)d] = n(4n+2)
=> 2a +(n-1)d = 8n + 4
=> 2a + (n-1) d = 12 - 8 + 8n
=> 2a + (n-1)d = 12 + 8( n-1)
On Comparing both sides, we get
a = 6
d = 8
Tn = a + (n-1)d
= 6 + (n-1) 8
= 6 + 8n - 8
= 8n - 2
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in RD answer is 4n-2
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Answered by
6
Sn = 4n² + 2n
=> ( first term ) a1 = S1 ( sum of first term )
• S1 = 4 + 2 = 6
a1 = 6
a2 = S2 - S1
• S2 = 4 × 4 + 4
S2 = 16 + 4
S2 = 20
a2 = 20 - 6
a2 = 14
• d = a2 - a1 ( common difference )
d = 8
an = a + ( n - 1 )d
an = 6 + ( n - 1 )8
an = 6 + 8n - 8
an = 8n - 2
an = 8n - 2
=> ( first term ) a1 = S1 ( sum of first term )
• S1 = 4 + 2 = 6
a1 = 6
a2 = S2 - S1
• S2 = 4 × 4 + 4
S2 = 16 + 4
S2 = 20
a2 = 20 - 6
a2 = 14
• d = a2 - a1 ( common difference )
d = 8
an = a + ( n - 1 )d
an = 6 + ( n - 1 )8
an = 6 + 8n - 8
an = 8n - 2
an = 8n - 2
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