if an AP of d= -5 n=125 a=4 then find the value of an
Answers
Step-by-step explanation:
Given that, a = 22, d = – 4 and Sn = 64 Let us consider the number of terms as n. For sum of terms in an A.P, we know that Sn = n2n2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So, ⟹ Sn = n2n2[2(22) + (n − 1)(−4)] ⟹ 64 = n2n2[2(22) + (n − 1)(−4)] ⟹ 64(2) = n(48 – 4n) ⟹ 128 = 48n – 4n2 After rearranging the terms, we have a quadratic equation 4n2 – 48n + 128 = 0, n2 – 12n + 32 = 0 [dividing by 4 on both sides] n2 – 12n + 32 = 0 Solving by factorisation method, n2 – 8n – 4n + 32 = 0 n ( n – 8 ) – 4 ( n – 8 ) = 0 (n – 8) (n – 4) = 0 So, we get n – 8 = 0 ⟹ n = 8 Or, n – 4 = 0 ⟹ n = 4 Hence, the number of terms can be either n = 4 or 8.
Answer:
-616
Step-by-step explanation:
an=a+(n-1) d
=4+(124)(-5)
=4-620
=-616