Question

Lesson:Arthematic Progression

1. 6th term of an Ap is -10 and 10th term is -26, find its 15th term.

correct and first answer will be marked as brainliest ​

Answer 1

━━━━━━━━━━━━━━━━━━━━

✤ Required Answer:

✒ GiveN:

• 6th term of an AP = -10
• 10th term of that AP = -26

✒ To FinD:

• Find its 15th term......?

━━━━━━━━━━━━━━━━━━━━

✤ How to Solve?

To solve this question, Let's know a little about AP

• An AP(Arithmetic progression) is a sequence of numbers where any two consecutive terms have equal difference. It means, Common difference is applicable to all consecutive terms.
• Terms of an AP are in the form: a, a + d, a + 2d, a + 3d......last term depends upon whether it is a finite or infinite series.

So, For this question, We just need to remember a formula for finding nth term,

$\large{ \boxed{ \sf{a_n = a + (n - 1)d}}}$

Where an is last term, a is first term, n is number of terms and d is common difference.

☕ So, Let's solve this question.

━━━━━━━━━━━━━━━━━━━━

✤ Solution:

According to formula,

• a6 = a + 5d
• a10 = a + 9d

We have, ATQ

• a6 = -10
• a10 = -26

So, Let's solve the equation,

➙ a6 = -10

➙ a + 5d = -10..........(1)

&

➙ a10 = -26

➙ a + 9d = -26...........(2)

Subtracting eq(1) from eq(2),

➙ a + 9d - (a + 5d) = -26 -(-10)

➙ a + 9d - a - 5d = -26 + 10

➙ 4d = -16

➙ d = -16/4 = -4

Putting value of d in eq(1),

➙ a + 5(-4) = -10

➙ a - 20 = -10

➙ a = 20 - 10

➙ a = 10

↪ Now, we have first term, common difference, and number of terms(Given ATQ)

So, By using formula,

➙ a15 = a + (15 - 1)d

➙ a15 = a + 14d

➙ a15 = 10 + 14(-4)

➙ a15 = 10 - 56

➙ a15 = -46

✒ Therefore, 15th term of the AP = -46

☀️ Hence, solved !!

━━━━━━━━━━━━━━━━━━━━

Answer 2

GIVEN :–

• 6th term of A.P. = -10

• 10th term of A.P. = -26

TO FIND :–

• 15th term = ?

SOLUTION :–

• We know that nth term –

$\\ \dashrightarrow \large { \boxed{ \bf T_{n} = a + (n - 1)d}}$

• According to the first condition –

$\\ \implies \large \bf T_{6} = a + (6- 1)d$

$\\ \implies \large \bf T_{6} = a + 5d$

$\\ \implies \large \bf - 10= a + 5d \:\:\: \: \: - - - - eq.(1)$

• According to the second condition –

$\\ \implies \large \bf T_{10} = a + (10- 1)d$

$\\ \implies \large \bf T_{6} = a + 9d$

$\\ \implies \large \bf -26= a + 9d \:\:\: \: \: - - - - eq.(2)$

• Subtract eq.(2) by eq.(1) –

$\\ \implies \large \bf - 10 - ( - 26)= a + 5d - a - 9d$

$\\ \implies \large \bf 16=- 4d$

$\\ \implies \large { \boxed{ \bf d = - 4 }}$

• Using eq.(1) –

$\\ \implies \large \bf - 10= a + 5(-4)$

$\\ \implies \large \bf - 10= a -20$

$\\ \implies \large { \boxed{ \bf a = 10}}$

• Now 15th term –

$\\ \implies \large \bf T_{15} = a + (15- 1)d$

$\\ \implies \large \bf T_{15} = a + 14d$

$\\ \implies \large \bf T_{15} = 10 + 14(-4)$

$\\ \implies \large \bf T_{15} = 10 -56$

$\\ \implies \large { \boxed{ \bf T_{15} = -46}}$