Question

Find the sum of 100 terms and 200 terms where the series 1, 3, 5, 7.​

Answer 1

First term = a₁ = 1

First term = a₁ = 1Second term = a₂ = 3

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 1

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = Sn

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = SnSn = n/2 [a + an]

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = SnSn = n/2 [a + an]     = 100/2 [ 1 + 199]

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = SnSn = n/2 [a + an]     = 100/2 [ 1 + 199]     = 50 [ 200]

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = SnSn = n/2 [a + an]     = 100/2 [ 1 + 199]     = 50 [ 200]     = 10000

First term = a₁ = 1Second term = a₂ = 3Common difference = d = a₂- a₁ = 3 - 1 = 2Last term = an = 199an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100No: of terms = n = 100Sum = SnSn = n/2 [a + an]     = 100/2 [ 1 + 199]     = 50 [ 200]     = 10000Sum = 10000

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Answer 2

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