2.The sum of all terms of the arithmetic progression having ten terms except for the first term is 99and except for the sixth term 89. Find the third term of the progression if the sum of the first termand the fifth term is equal to 10
(a) 15
(b) 5
(c) 8
(d) 10
Answers
Answer:
a3 = 5
Step-by-step explanation:
The formula used will be -
Sn = n/2[2a+(n-1)d]
S10 = 5[2a+(9)d]
S1= 1/2{2a]=>a
5[2a+9d]-a=99
5[2a+9d]-a=99
9a+45d=99
/9
a+5d=11 (1 eq)
a+2d = 5 (2 eq)
5[2a+9d]-11=89
10a+45d=100 (3 eq)
Solving (1) & (3)
a=1, d=2
Thus,
a3=1+4
a3=5
Given,
Sum of all terms having ten terms except for the first term is 99 and except for the sixth term 89. Sum of 1st and 5th term is equal to 10.
To Find,
Third term of the progression.
Solution,
First term is a and common difference is d.
Sum of first and third terms is 10.
t₁ + t₅ = 10
==> a + (1 - 1) * d + a + (5 - 1) * d = 10
==> a + a + 4d = 10
==> 2a + 4d = 10
==> a + 2d = 5 ---- (1)
Ten Terms except for the first term is 99 and except for sixth term is 89.
S₁₀ - First term = 99
==> 10/2[2a + (10 - 1) * d] - a = 99
==> 5[2a + 9d] - a = 99
==> 10a + 45d - a = 99
==> 9a + 45d = 99
==> a + 5d = 11 ------- (2)
Subtract Equation (1) and (2).
a + 5d - a - 2d = 11 - 5
==> 3d = 6
==> d = 2
Place d = 2 in (2).
==> a + 5d = 11
==> a + 10 = 11
==> a = 1
Third term of the progression.
t₃ = 1 + (3 - 1) * 2
t₃ = 5
Result,
Third term of the progression is 5.