the first term of an AP is 5 the last term is 45 and the sum of its terms is 1000. find the number of terms and common difference of the AP
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nth term (an) = a+(n-1)d
a₁ =5 =a
an = 45
45 = 5+(n-1)d
40 = (n-1)d ⇒ nd - d = 40 →(1)
Sn = 1000
Sn = n/2 (2a+(n-1)d)
1000 = n/2 (2*5+(n-1)d)
2000 = n(10+nd-d)
substituting "nd-d" as 40 we get,
2000 = n(10+40)
2000 = 50n
⇒ n= 40
substituting the value in (1)
nd-d = 40
d(n-1) = 40
d(40-1)= 40
d = 40/39
Hope it helps..... :)
a₁ =5 =a
an = 45
45 = 5+(n-1)d
40 = (n-1)d ⇒ nd - d = 40 →(1)
Sn = 1000
Sn = n/2 (2a+(n-1)d)
1000 = n/2 (2*5+(n-1)d)
2000 = n(10+nd-d)
substituting "nd-d" as 40 we get,
2000 = n(10+40)
2000 = 50n
⇒ n= 40
substituting the value in (1)
nd-d = 40
d(n-1) = 40
d(40-1)= 40
d = 40/39
Hope it helps..... :)
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