The nth term of an AP is given by Tn=3n+5. Find the sum of the first forty terms.
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Answer:nth Term of A. P.
Tn = 3n + 4
Putting the value of n = 1, 2, 3.
For n = 1,
T1 = 3 *1 + 4 = 3 + 4 = 7
For n = 2,
T2 = 3 *2 + 4 = 6 + 4 = 10
For n = 3,
T3 = 3 *3 + 4 = 9 + 4 = 13.
T2 - T1 = d
10 - 7 = 3
d = 3.
T3 - T2 = d
13 - 10 = 3.
Since, difference is common.
So, ✔️ Common difference = 3.
♠️AP. = 7, 10, 13.....
SUM OF FIRST FIVE TERMS
=> Sn = n / 2 [ 2a + ( n - 1 ) d ]
=> S5 = 5 / 2 [ 2 *7 + ( 5 - 1 ) 3 ]
=> S5 = 5 / 2 [ 14 + 4*3 ]
=> S5 = 5 / 2 [ 14 + 12 ]
=> S5 = 5 / 2 [ 26]
=> S5 = 5 * 13
=> S5 = 65.
Sum of first five terms = 65.
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