Question

find the sum of first 17 term of ap whose nth term is given by Tn=7-4n

Answer 1

Heya !!!


Tn = 7 - 4N


T1 = 7 - 4 × 1




=> 7-4



=> 3




T2 = 7 - 4 × 2





=> 7 - 8




=> -1



And,




T3 = 7 - 4 × 3




=> 7-12




=> -5



First term (A) = 3


And,


Common Difference (D) = -1-3 = -4




Therefore,


Sn = N/2 × [ 2A + (N-1) × D ]






S17 = 17/2 × [ 2 × 3 + (17-1) × -4 ]







=> 17/2 ×{ ( 6 + (-64) }



=> 17/2 × (6-64)





=> 17/2 × -58






=> 17 × 29





=> 493



Hence,




Sum of first 17 terms of an AP is 493.



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Answer 2

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tn = 7-4n

t1 = 7-4×1 = 3

t2 = 7-4×2 = - 1

t3 = 7-4×3 = - 5

Common difference (d) =-1-3 =-4

a = 3 n = 17

$sn = \frac{n}{2} \times ( 2a + (n - 1)d)$

$S17 = \frac{17}{2}(2 \times 3 + 16 \times - 4)$

$S17 = \frac{17}{2} \times (6 - 64)$

$S17 = \frac{17}{2} \times - 58$

$S17 = 17 \times - 29$

$S17 = - 493$

Therefore, the sum of the first 17 terms of the AP is - 493.