Find the sum of 12 terms of an AP -7, -3, 1 starting from the eight term
Answers
Answered by
1
an = a+(n-1)d
a12 = -7+(11)4
a12 = 37
ATQ
Sn = n/2{a+(n-1)d}
S12 = 6{37+(11)4}
S12 = 486
a12 = -7+(11)4
a12 = 37
ATQ
Sn = n/2{a+(n-1)d}
S12 = 6{37+(11)4}
S12 = 486
Answered by
0
a = t1 =-7
common difference (d) = t2- t1
= -3 - (-7)
= -3 +7
= 4
s12 = n÷2 [2a+(n-1) × d ]
s12 = 12÷2 [ 2×-7 + (12-1) × 4]
= 6 [ -14 + 11×4 ]
= 6 (- 14× 44)
= 6 × 616
s12 = 3696
common difference (d) = t2- t1
= -3 - (-7)
= -3 +7
= 4
s12 = n÷2 [2a+(n-1) × d ]
s12 = 12÷2 [ 2×-7 + (12-1) × 4]
= 6 [ -14 + 11×4 ]
= 6 (- 14× 44)
= 6 × 616
s12 = 3696
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