if the third term of an ap is 1 and 6 th term is- 11 find the sum of first 32 twrms
Answers
In an A.P [Arithmetic Progression]
- aₙ = a + (n - 1)d
→ a₃ = a + 2d
→ 1 = a + 2d ___(1)
→ a₆ = a + 5d
→ - 11 = a + 5d ___(2)
By doing (1) - (2),
→ 12 = - 3d
→ d = - 4
Then, a = 1 - 2d
→ 1 - 2(- 4)
→ 9
Sₙ = n/2 [2a + (n - 1)d]
→ S₃₂ = 32/2 [2(9) + (32 - 1)(- 4)]
→ S₃₂ = 16 [18 - 124]
→ S₃₂ = 16 (- 106)
→ S₃₂ = - 1696
The sum of 32 terms in A.P are - - 1696.
Answer:
The sum of 32 terms in an A.P. are -53.
Step-by-step explanation:
Given,
Third term, a3 = 1
Sixth term, a6 = -11
By using the formula of A.P.
→an = a + (n - 1)d
=> a3 = a + 2d
=> 1 = a + 2d ----------(1)
=> a6 = a + 5d
=> -11 = a + 5d ------(2)
On subtracting equation (1) and (2),
=> 12 = -3d
=> 12/-3 = d
=> -4 = d
=> d = -4
.°. d = -4
Now,
=> a = 1 - 2d
Putting the given values.
=> a = 1 - 2 (-4) [°.° (-) × (-) = (+) ]
=> a = 1 + 8
=> a = 9
.°. a = 9
Now,
Sn = n/2 [ 2a + (n - 1)d ]
=> S32 = 32/2 [ 2 (9) + (32 - 1)(-4) ]
=> S32 = 32/2 [ 18 - 124 ]
=> S32 = 32/2 × (-106)
=> S32 = -1,696
.°. S32 = -1,696