What is the 32nd term of the arithmetic sequence where a1 = 14 and a13 = -58?

What is the 32nd term of the arithmetic sequence where a1 = 14 and a13 = -58?

=D

The correct option is (B) -172. Step-by-step explanation:  We are given to find the 32nd term of the arithmetic sequence where the 1st term and 13th term are We know that the n-th term of an arithmetic sequence with first term and common difference is given by According to the given information, we have and Therefore, the 32nd term will be Thus, the 32nd term is -172. Option (B) is correct.

(-120 – (-32))/(9 – 1) = -11 a(n) = -32 – 11*(n – 1) a(32) = -32 – 11*(32 – 1) = -373

Answer 6

(-57 – 15)/(13 – 1) = -6 a(n) = 15 – 6·(n – 1) a(32) = 15 – 6·(32 – 1) = -171

Answer 7

– 174 Step-by-step explanation: The nth term of an arithmetic  sequence is given as Tn = a + (n – 1)d where Tn is the nth term a is the first term , n is the number of term and d is the common difference. As such, a13 = a + (13 – 1)d = a + 12d Given that a1 = 12 and a13 = -60 -60 = 12 + 12d 12d = -72 d = -6 Hence a32 which is the 32nd term = 12 + (32 – 1)-6 = 12 + (-186) = – 174

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-373 Step-by-step explanation: A n = A 1 + d ( n − 1 ) A 1 = − 32 A 9 = − 120 ⇒ A 9 = − 32 + d ( 9 − 1 ) ⇒ − 120 = − 32 + d ( 8 ) ⇒ − 88 = 8 d ⇒ d = − 11 A 32 = − 32 + ( − 11 ) ( 32 − 1 ) ⇒ A 32 = − 32 + ( − 11 ) ( 31 ) ⇒ A 32 = − 32 + − 341 ⇒ A 32 = − 373

Step-by-step explanation: Given : the arithmetic sequence where and We have to find the 32nd term of the arithmetic sequence. Since, the general arithmetic sequence having first term ‘a’ and common difference ‘d’ is given by Thus, for the given arithmetic sequence, we have, First term is and Calculate the common difference by putting a = 13 in above, we have, Solving for d, we have, Divide by 12 both side, we have, Thus, the common difference is -6. For 32nd term, Put a = 13 , d = -6 and n = 32 in We have, Simplify, we have,

a 32 = -171 Step-by-step explanation: a.-171

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