The sum of all the 10 terms of an AP is 485.if is last term is 71, find the first term and the common difference of the AP
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Given,
S10 = 485
an = 71
Find - a and d
S10 = n/2[a+l]
485 = 10/2 [ a + 71 ]
485 = 5 [ a + 71 ]
485/5 = a+ 71
97 - 71 = a
a = 26
Also,
S10 = n/2 [ 2a + ( n-1 )d ]
485 = 10/2 [ 2(26) + (10-1) d ]
485 = 5[ 52+10d ]
485/5 = 52 + 10d
97 - 52 = 10d
45/10 = d
d = 4.5
Hence solved
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S10 = 485
an = 71
Find - a and d
S10 = n/2[a+l]
485 = 10/2 [ a + 71 ]
485 = 5 [ a + 71 ]
485/5 = a+ 71
97 - 71 = a
a = 26
Also,
S10 = n/2 [ 2a + ( n-1 )d ]
485 = 10/2 [ 2(26) + (10-1) d ]
485 = 5[ 52+10d ]
485/5 = 52 + 10d
97 - 52 = 10d
45/10 = d
d = 4.5
Hence solved
Hope it helps you mark it brainlist its free
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