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The first three terms of an arithmetic sequence are −5, k, 1 - Leaving Cert Mathematics - Question 6 - 2021

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The first three terms of an arithmetic sequence are −5, k, 1. (i) Find k and hence show that the common difference is 3. (ii) Find T_{10}, the 10th term in the se... show full transcript

Worked Solution & Example Answer:The first three terms of an arithmetic sequence are −5, k, 1 - Leaving Cert Mathematics - Question 6 - 2021

Step 1

Find k and hence show that the common difference is 3.

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Answer

In an arithmetic sequence, the difference between consecutive terms is constant. We can express this relationship as:

T2T1=T3T2T_2 - T_1 = T_3 - T_2 Substituting the given terms:

k(5)=1kk - (-5) = 1 - k This simplifies to: k+5=1kk + 5 = 1 - k Adding k to both sides gives: 2k+5=12k + 5 = 1 Next, subtract 5 from both sides: 2k=152k = 1 - 5 Which simplifies further to: 2k=42k = -4 Hence, we find: k=2k = -2

Now, to find the common difference: The first term is -5 and the second term is -2.

Common difference: d=2(5)=2+5=3d = -2 - (-5) = -2 + 5 = 3 Thus, the common difference is confirmed as 3.

Step 2

Find T_{10}, the 10th term in the sequence.

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Answer

Using the formula for the nth term of an arithmetic sequence:

Tn=T1+(n1)dT_n = T_1 + (n - 1)d

For this sequence, the first term, T1=5T_1 = -5, and the common difference d=3d = 3.

To find T10T_{10}:

T10=5+(101)3T_{10} = -5 + (10 - 1)3 T10=5+9=4T_{10} = -5 + 9 = 4 Therefore, T10=22T_{10} = 22.

Step 3

Find which term in the sequence has a value of 247.

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Answer

We start again with our nth term formula:

Tn=T1+(n1)dT_n = T_1 + (n - 1)d

We want to find n such that:

Tn=247T_n = 247 Substituting what we know:

247=5+(n1)3247 = -5 + (n - 1)3 Adding 5 to both sides:

252=(n1)3252 = (n - 1)3 Dividing both sides by 3:

n1=84n - 1 = 84 Therefore: n=85n = 85 Thus, the 85th term in the sequence has a value of 247.

Step 4

Find S_{50}, the sum of the first 50 terms of the sequence.

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Answer

To find the sum of the first n terms of an arithmetic sequence, we use the formula:

Sn=n2(2a+(n1)d)S_n = \frac{n}{2} (2a + (n - 1)d) Where:

  • n=50n = 50 (number of terms)
  • a=4a = 4 (first term)
  • d=5d = 5 (common difference found from 4 to 9 and 9 to 14)

Plugging the values into the formula:

S50=502(24+(501)5)S_{50} = \frac{50}{2} (2 \cdot 4 + (50 - 1)5) S50=25(8+245)S_{50} = 25 (8 + 245) S50=25253S_{50} = 25 \cdot 253 Therefore, performing the multiplication: S50=6325S_{50} = 6325

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