The sum of first four terms of an A.P. is 56 and
the sum of its last four terms is 112. If its first
term is 11, then number of its terms is-
(A) 10
(B) 11
(C) 12
(D) None of these
plz give answer with method
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The sum of first four terms of an A.P. is 56 and
the sum of its last four terms is 112. If its first
term is 11, then number of its terms is-
(A) 10
(B) 11
(C) 12
(D) None of these
plz give answer with method
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Answer:
A)10
Step-by-step explanation:
SUM OF FIRST FOUR TERMS=a+(a+d)+(a+2d)+(a+3d)=4a+6d=56
and FIRST TERM,a=11 SUBSTITUTING IN ABOVE EQUATION WE GET , d=2.
sum of last four terms=(a+nd)+(a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)=4a+4nd-6d=112
by substituting value of a=11 and d=2 in above equation , we get n=10
i.e., number of terms,n=10.
Answer:
(B). 11
Step-by-step explanation:
let the first four terms are a, a+d, a+2d, a+3d
therefore,
a+a+d+a+2d+a+3d = 56
a = 11 given
4a +6d = 56
44 + 6d =56
6d = 12
d=2
let there are n terms in the AP
therefore,
last four terms are,
a+(n-1)d, a+(n-2)d, a+(n-3)d, a+(n-3)d
a+(n-1)d + a+(n-2)d, + a+(n-3)d + a+(n-4)d = 4a + 4nd - 10d = 112 given
4×11 + 4n×2 - 10×2 = 112
44 + 8n -20 = 112
8n = 88
n = 11