The value of a28 given a11 = 23 and a15 = −5 is -96
Given:
a11 = 23
a15 = −5
Where,
a1 = first term
d = common difference
a11 = a + 10d = 23
a15 = a + 14d = -5
a + 10d = 23 (1)
a + 10d = 23 (1)a + 14d = -5 (2)
subtract (1) from (2) to eliminate a
14d - 10d = -5 - 23
4d = -28
d = -28 / 4
d = -7
Substitute d = -7 into (1)
a + 10d = 23 (1)
a + 10(-7) = 23
a - 70 = 23
a = 23 + 70
a = 93
So,
a28 = a + 27d
= 93 + 27(-7)
= 93 + -189
= 93 - 189
= -96
Therefore, the value of a28 given a11 = 23 and a15 = −5 is -96
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