Question 6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Class X1 - Maths -Sequences and Series Page 199
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A/C to question,
series is 13, 17, 21, ......97
first term (a) = 13
Common difference (d) = 17 - 13 = 4
nth term ( Tn) = 97
use the formula,
Tn = a + (n - 1)d
97 = 13 + (n-1)4
97 - 13 = 4(n - 1)
84 = 4(n - 1)
21 = n - 1
n = 22
hence, no of terms = 22
now,
use formula ,
sum of n terms ( Sn) = n/2{ 2a + (n-1)d }
S₂₂ = 22/2 { 2 × 13 + ( 22 - 1) × 4 }
= 11 × { 26 + 21 × 4 }
= 11 × { 26 + 84 }
= 11 × 110
= 1210
series is 13, 17, 21, ......97
first term (a) = 13
Common difference (d) = 17 - 13 = 4
nth term ( Tn) = 97
use the formula,
Tn = a + (n - 1)d
97 = 13 + (n-1)4
97 - 13 = 4(n - 1)
84 = 4(n - 1)
21 = n - 1
n = 22
hence, no of terms = 22
now,
use formula ,
sum of n terms ( Sn) = n/2{ 2a + (n-1)d }
S₂₂ = 22/2 { 2 × 13 + ( 22 - 1) × 4 }
= 11 × { 26 + 21 × 4 }
= 11 × { 26 + 84 }
= 11 × 110
= 1210
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