t1+t5+t10+t15------=225 find the sum of first 24 terms of AP
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Question:-)S24 = 24/2 [ 2A + (24-1)D]
= 24A + 276D
The sum of the first, 5th, 10th, 15th, 20th and 24th term = 225
A + ( A + 4D) +( A + 9D) + ( A + 14D) + ( A + 19D) + ( A + 23D)= 225
; 6A + 69D= 225 - (1)
Multiply (1) by 6 and you get
24A + 276D = 1350.
Answer:-)S24 = 24/2 [ 2A + (24-1)D]
= 24A + 276D
The sum of the first, 5th, 10th, 15th, 20th and 24th term = 225
A + ( A + 4D) +( A + 9D) + ( A + 14D) + ( A + 19D) + ( A + 23D)= 225
; 6A + 69D= 225 - (1)
Multiply (1) by 6 and you get
24A + 276D = 1350.
= 24A + 276D
The sum of the first, 5th, 10th, 15th, 20th and 24th term = 225
A + ( A + 4D) +( A + 9D) + ( A + 14D) + ( A + 19D) + ( A + 23D)= 225
; 6A + 69D= 225 - (1)
Multiply (1) by 6 and you get
24A + 276D = 1350.
Answer:-)S24 = 24/2 [ 2A + (24-1)D]
= 24A + 276D
The sum of the first, 5th, 10th, 15th, 20th and 24th term = 225
A + ( A + 4D) +( A + 9D) + ( A + 14D) + ( A + 19D) + ( A + 23D)= 225
; 6A + 69D= 225 - (1)
Multiply (1) by 6 and you get
24A + 276D = 1350.
Answered by
2
Answer-
S24th term= 24/2(2a+(24-1)d)
=12(2a+23d)
=24a+276d........(i)
NOW
Tn= a+(n-1)d
SO....
t1+t5+t10+t15+t20+t24=(a+a+4d+a+9d+a+14d+a+19d+a+23d)
= 6a+69d=225.......(ii)
multiply eq (ii) by 4
24a+276d=900..
answer will be 900..
Step-by-step explanation:
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