if the 8th term of an AP is 31 and the 15 is 16 more then the 11th term find the AP
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Answered by
32
Step-by-step explanation:
Let a be the 1st term and d be the common difference
a8 = a + 7d = 31
a15 = a + 14d
= a11 + 16(a + 10d) + 16a + 10d +16
= a + 14d = a + 10d +16 = a + 10d + 4d= 16 = 4d
a8 = a + 7d = 31 a15 = a + 14d = a11 + 16(a + 10d) + 16a + 10d +16 = a + 14d = a + 10d +16 = a + 10d + 4d
= 16 = 4d
d = 16/4 = 4
a + 7d = 31
a + 7d = 31a + 7(4) = 31
a + 7d = 31a + 7(4) = 31a + 28 = 31 a = 31 -28 = 3a = 3 , d = 4 ...
therefore Ap = a , a+d , a+2d ,a+3d ......= 3 , 3+4 ,3+2(4) , 3 + 3(4)= 3 , 7 , 11 , 15 .......
Answered by
5
Answer:
A8=31
=> a + 7d=31...... 1
given
a15=16 + a11
=> a + 14d = a + 10d + 16
=> 4d = 16
=> d = 4
now, put the value of d in equation 1
a + 7(4) = 31
a = 31 - 28
a = 3
here a = 3
d = 4
Ap is--- 3,7,11......
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