Find an equation for the nth term of the arithmetic sequence.
a10 = 32, a12 = 106
Solution:
The nth for the arithmetic sequence is given by:
an = a + (n - 1)d --- (1)
Where, a is the first term
d is the common difference of two consecutive terms.
n is the number of terms.
It is given that:
a10 = 32 and a12 = 105
From equation 1 we have,
⇒ a10 = a + 9d
⇒ a + 9d = 32 --- (2)
⇒ a12 = a + 11d
⇒ a + 11d = 106 --- (3)
Subtract equation [2] from [3] we have;
⇒ a + 11d - a - 9d = 106 - 32
By simplification we get,
⇒ 2d = 74
By dividing both sides by 2 we get,
⇒ d = 37
Substitute the value of d in [2] we have;
⇒ a + 9(37) = 32
⇒ a + 333 = 32
Subtract 333 from both sides we have;
⇒ a = -301
Then substitute the value a and d in equation [1] we have;
⇒ an = - 301 + (n - 1)(37)
⇒ an = - 301 + 37n - 37
⇒ an = - 338 + 37n
Therefore, an = - 338 + 37n an equation for the nth term.
Find an equation for the nth term of the arithmetic sequence.
a10 = 32, a12 = 106
Summary:
an = - 338 + 37n an equation for the nth term of the arithmetic sequence. a10 = 32, a12 = 106.
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