OR
In an AP of 50 terms, the sum of first 10 terms is 210 and sum of last 15 terms is 2565,
find AP
Answers
a = first term
d = common difference
n = terms
★ Sum of first 10 terms of AP is 210 [Given]
Sn = n/2[2a + (n - 1)d]
➠10/2 [ 2a + (10 - 1)d] = 210
➠5[2a + 9d] = 210
➠10a + 45d = 210......( i )
★Sum of last 15 terms is 2565 and AP has 50 terms [Given]
S50 - S35 = 2565
➠50/2 [ 2a + (50 - 1)d] - 35/2 [ 2a + (35 - 1)d] = 2565
➠25[2a + 49d] - 35/2 [ 2a + 34d] = 2565
➠50a + 1225d - 35a - 595d = 2565
➠15a + 630d = 2565
➠a + 42d = 171 ....................( ii )
From (i) and (ii)
10a + 45d = 210
10a + 420d = 1710
-------------------------------
- 375d = -1500
⟼d = 1500/375
⟼d = 4
✪Putting the value of d in (ii)
a + 42(4) = 171
a + 168 = 171
a = 171 - 168
a = 3
Now,
a = 3
a + d = 3 + 4 = 7
a + 2d = 3 + 4(2) = 3 + 8 = 11
a + 3d = 3 + 4(3) = 3 + 12 = 15
Hence, AP is 3, 7, 11, 15.....
a = first terms
d = common different
n = terms
Sum of first 10 terms of Ap is 210
Sn = n/2[2a + (n - 1)d]
= 10/2[2a + (10 - 1)d]
5[2a + 9d] = 210
10a + 45d = 210.......(1)
Sum of last 15 terms is 2565 and Ap has 50 terms. [Given]
S50 - S35 = 2565
50/2[2a + (50 - 1)d] - 35/2[2a + (35 - 1)d] = 2565
25[2a + 49d] - 35/2[2a + 34d] = 2565
50a + 1225d - 35a - 595d = 2565
15a + 630d = 2565
a + 42d = 171........(2)
from (1) and (2)
(10a + 45d = 210) - (10a + 420d =1710) = -375d = -1500
d = 1500/375
d = 4
putting the value of d in (2)
a + 42(4) = 171
a + 168 = 171
a = 3
Now,
a = 3
a + d = 3 + 4 = 7
a + 2d = 3 +4(2) = 3 + 8 = 11
a + 3d = 3 + 4(3) = 3 + 12 = 15
Hence, Ap is 3,7,11,15...
silentlover45.❤️