Find the 37th term of an arithmetic sequence whose second and third terms are −4 and 12. If the fourth term of an arithmetic sequence is 17 and the second term is 3, find the 24th term.

Question

Answer 1

Step 1 Given: Second term

(a_{2}) = - 4

Third term

(a_{3}) = 12

Step 2 Used concept

T_{n} = a_{1} + (n - 1) d

Where

T_{n} \to n^{(t h)} term

a_{1} \to 1^{(s t)} term

d \to difference = (a_{2} - a_{1}) = (a_{3} - a_{2})

Step 3 Apply the above concept it gives

d = a_{3} - a_{2}

d = 12 - (- 4)

d = 12 + 4 = 16

now,

a_{1} = a_{2} - d

a_{1} = - 4 - 16 = - 20

Step 4 The

37^{(t h)}

term of the given arithmetic sequence will be

T_{37} = - 20 + (37 - 1) \times 16

= - 20 + 36 \times 16

= - 20 + 576

= 556

(answer)

Find the 37th term of an arithmetic sequence whose second and third terms are −4 and 12. If the fourth term of an arithmetic sequence is 17 and the second term is 3, find the 24th term.

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