Question

In an arithmetic sequence

S10 = 125, a = 17 find the common difference, d,of the sequence
please answer fast please​

Answer 1

Answer:

nth term of AP, tn = a + (n-1) d

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common difference

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and d

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)(1)*2 ⇒ 2a + 4d = 30 ---(3)

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)(1)*2 ⇒ 2a + 4d = 30 ---(3)(2) - (3) ⇒

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)(1)*2 ⇒ 2a + 4d = 30 ---(3)(2) - (3) ⇒5d = -5

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)(1)*2 ⇒ 2a + 4d = 30 ---(3)(2) - (3) ⇒5d = -5d = -1

nth term of AP, tn = a + (n-1) d Sum of n terms Sn = n/2[2a + (n-1)d]a - first term and d = common differenceTo find a and dIt is given that, a3=15 ,S10=125We can write a + 2d = 15 ----(1)10/2[2a + 9d ] = 125⇒5[2a + 9d ] = 125⇒2a + 9d = 25 ----(2)(1)*2 ⇒ 2a + 4d = 30 ---(3)(2) - (3) ⇒5d = -5d = -1eq (1) ⇒ a + 2d = 15

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