Question

The 6th &17th term of an AP are 19 & 41 respectively. Find the 40th term

Answer 1

Answer:

a40 = 767

Step-by-step explanation:

Let first term = a

common difference = d

==: a6 = 19

==: 19 = a + (6 - 1)d

==: 19 = a + 5d

==: a + 5d = 19        ...............(1)

==: a17 = 41

==: 41 = a + (17 - 1)d

==: 41 = a + 16d

==: a + 6d = 41        ...............(2)

From 1 and 2 we get

a = -91

d = 22

a40 = a + (n - 1) d

a40 = -91 + (40 - 1) 22

a40 = -91 + 880 - 22

a40 = 767

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Answer 2

Given:-

• The 6th &17th term of an AP are 19 & 41 respectively.

To Find:-

• Find the 40th term..?

Solutions:-

• 6th term of Ap is 19
• 17th term of Ap is 41

We know that:-

The 6th term of Ap is 19

=> an = a + (n - 1)d

=> a6 = a + (6 - 1)d

=> 19 = a + 5d ...........(i).

The 17th term of Ap is 41

=> an = a + (n - 1)d

=> a17 = a + (17 - 1)d

=> 41 = a + 16d ...........(ii).

Now, Subtracting Eq. (ii) and (i) we get,

${a} \: + \: {16d} \: \: = \: \: {41} \\ {a} \: + \: {5d} \: \: = \: \: {19} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: {11d} \: \: \: \: \: \: \: \: = \: \: \: {22}$

=> d = 22/11

=> d = 2

Now, putting the value of y in Eq. (i).

=> a + 5d = 19

=> a + 5(2) = 19

=> a + 10 = 19

=> a = 19 - 10

=> a = 9

Thus, a = 9, d = 2 and n = 40

=> an = a + (n - 1)d

=> a40 = 9 + (40 - 1) (2)

=> a40 = 9 + 39 × 2

=> a40 = 9 + 78

=> a40 = 87

Hence, the 40th term is 87.