Question

plz plz answer fast by formula of Sn

Answer 1

last term = Tn = 803
first term = a = 3
common difference = d = T2-T1 = 11-3 = 8
Sum of terms = Sn = ?
number of terms = n = ?

Tn = a+(n-1)d
803 = 3+(n-1)8
803-3 = 8n-8
800+8 = 8n
808/8 = n
101 = n

Sn = n/2[a+Tn]
S101 = 101/2[3+803]
S101 = 101/2[806]
S101 = 101 [403]
S101 = 40703

so the sum of the A.P is 40703

Answer 2

$Hey\:There$ ‼
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A.P. 3 + 11 + 19 + ... + 803

First term, a = 3

Common distance, d = a2 – a1
= 11 – 3
= 8

Last term, aⁿ = 803

As we know,
aⁿ = a + ( n – 1 ) d
803 = 3 + ( n – 1 ) 8
803 – 3 = ( n – 1 ) 8
800 = ( n – 1 ) 8
800 / 8 = n – 1
100 = n – 1
n = 100 + 1
n = 101

Sum, S = n/2 [ 2a + ( n – 1 ) d ]
=> 101 / 2 × [ 2 × 3 + ( 101 – 1 ) 8 ]
=> 101 / 2 × [ 6 + 100 × 8 ]
=> 101 / 2 × [ 6 + 800 ]
=> 101 / 2 × 806
=> 101 × 403
=> 40703
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