3. The 3rd term of an arithmetic sequence is 58 and the 10th term is 30. Find a1...

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aexxiren @aexxiren

September 2022 1 1 Report

3. The 3rd term of an arithmetic sequence is 58 and the 10th term is 30. Find a1, d and the nth term.
with solution

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Solution:

Step 1: Equation for 3rd term

$a_{n}=a_{1}+(n-1)d\\a_{3}=a_{1}+(n-1)d\\58=a_{1}+(3-1)d\\58=a_{1}+2d\\a_{1}=-2d+58$

Step 2: Equation for 10th term

$a_{n}=a_{1}+(n-1)d\\a_{10}=a_{1}+(n-1)d\\30=a_{1}+(10-1)d\\30=a_{1}+9d\\a_{1}=-9d+30$

Step 3: Solve for $d$

$-2d+58=-9d+30\\-2d+9d=30-58\\7d=-28\\d=-4$

Step 4: Solve for $a_{1}$ (using step 1 equation)

$a_{1}=-2d+58\\a_{1}=-2(-4)+58\\a_{1}=8+58\\a_{1}=66$

ImNobody64 7d = -28 ... d = 4

ImNobody64 how?

raaffy divide both sides (of equation) by 7

raaffy 7d / 7 = -28 / 7 ... d = -4

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