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Answer 1

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Let first term of AP = a

Let first term of AP = acommon difference = d

Let first term of AP = acommon difference = dnumber of terms = n

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49=> (7/2)*{2a + (7-1)*d} = 49

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49=> (7/2)*{2a + (7-1)*d} = 49 => (7/2)*{2a + 6d} = 49

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49=> (7/2)*{2a + (7-1)*d} = 49 => (7/2)*{2a + 6d} = 49=> (1/2)*{2a + 6d} = 7

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49=> (7/2)*{2a + (7-1)*d} = 49 => (7/2)*{2a + 6d} = 49=> (1/2)*{2a + 6d} = 7=> (2/2)*{a + 3d} = 7

Let first term of AP = acommon difference = dnumber of terms = nGiven that sum of first 7 term of AP = 49=> (n/2)*{2a + (n-1)*d} = 49=> (7/2)*{2a + (7-1)*d} = 49 => (7/2)*{2a + 6d} = 49=> (1/2)*{2a + 6d} = 7=> (2/2)*{a + 3d} = 7=> a + 3d = 7 ...1

...1Again given sum of first 17 term of AP = 289

...1Again given sum of first 17 term of AP = 289=> (17/2)*{2a + (17-1)*d} = 289

...1Again given sum of first 17 term of AP = 289=> (17/2)*{2a + (17-1)*d} = 289=> (1/2)*{2a + 16*d} = 17 (when 17 and 289 is divided by 17)

...1Again given sum of first 17 term of AP = 289=> (17/2)*{2a + (17-1)*d} = 289=> (1/2)*{2a + 16*d} = 17 (when 17 and 289 is divided by 17)=> (2/2)*{a + 8d} = 17

...1Again given sum of first 17 term of AP = 289=> (17/2)*{2a + (17-1)*d} = 289=> (1/2)*{2a + 16*d} = 17 (when 17 and 289 is divided by 17)=> (2/2)*{a + 8d} = 17=> a + 8d = 17 ...2

...2After solving equation 1 and 2, we get

...2After solving equation 1 and 2, we geta = 1 and d = 2

...2After solving equation 1 and 2, we geta = 1 and d = 2Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}

...2After solving equation 1 and 2, we geta = 1 and d = 2Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}=> (n/2)* {2*1 + (n-1)*2}

...2After solving equation 1 and 2, we geta = 1 and d = 2Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}=> (n/2)* {2*1 + (n-1)*2}=> (n/2)* (2 + 2n - 2)

...2After solving equation 1 and 2, we geta = 1 and d = 2Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}=> (n/2)* {2*1 + (n-1)*2}=> (n/2)* (2 + 2n - 2)=> (n/2)*2n

...2After solving equation 1 and 2, we geta = 1 and d = 2Now sum of n terms of AP = (n/2)*{2a + (n-1)*d}=> (n/2)* {2*1 + (n-1)*2}=> (n/2)* (2 + 2n - 2)=> (n/2)*2n=> n2

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Answer 2

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