the sum of first m terms of an A.P. is 4m^2-m. If its nth term is 107 find the value of n. Also find the 21st term of this A.P.?
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Here devil1407
Given:
Sum of first m terms of an A.P = 4m² - m
m - 1 , ... S1 = 4×1²-1 = 3
S1 = a ⇒ a = 3
m - 2 , ....S2 = 4×2² - 2 = 16-2 = 14
S2 = a1 + a2 = a + a + d = 14
⇒ 2a + d = 14
2 × 3 + d = 14
6 + d =14
d = 14 - 6
d = 8
Given:
an = 107
a +(n-1) d = 107
3 + (n-1)8 = 107
3 + (n-1)8 = 107
(n-1)8 = 107 - 3
8n - 8 = 104
8n = 104 +8
n = 112/8
n = 14
Find a21:
a21 = 3 + (21 - 1 ) 8
= 3 + 20 × 8
= 3 + 160
a21 = 163
Hope it helps U
Thanx...Bye
Here devil1407
Given:
Sum of first m terms of an A.P = 4m² - m
m - 1 , ... S1 = 4×1²-1 = 3
S1 = a ⇒ a = 3
m - 2 , ....S2 = 4×2² - 2 = 16-2 = 14
S2 = a1 + a2 = a + a + d = 14
⇒ 2a + d = 14
2 × 3 + d = 14
6 + d =14
d = 14 - 6
d = 8
Given:
an = 107
a +(n-1) d = 107
3 + (n-1)8 = 107
3 + (n-1)8 = 107
(n-1)8 = 107 - 3
8n - 8 = 104
8n = 104 +8
n = 112/8
n = 14
Find a21:
a21 = 3 + (21 - 1 ) 8
= 3 + 20 × 8
= 3 + 160
a21 = 163
Hope it helps U
Thanx...Bye
ketharin:
Nee mudichitiya
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