Question

find the 40th term of the arithmetic sequence whose 9 th term is 465 and 20th term is 368 ​

Answer 1

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tn = a + (n – 1) d t9 = a + 8d (t9 = 465) a + 8d = 465 … (1) Substitute the value of d = -7 in (1) a + 8(-7) = 465 a – 56 = 465 a = 465 + 56 = 521 a = 521, d = -7, n = 40 t40 = 521 + 39(-7) = 521 – 273 = 248 40th term of an A.P. is 248

Answer 2

Step-by-step explanation:

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