Question

which term of the A.P 4,11,18,25,....42 more than its 25th term?​

Answer 1

Given:-

• A.P. = 4, 11, 18, 25, ....
• a (first term) = 4
• d (common difference) = 11 - 4 = 7

To Find:-

• Which term of the A.P. is 42 more than 25th term.

Solution:-

Firstly we'll find the 25th term of the given A.P.

We have:-

• a = 4
• d = 7
• n = 25

We know,

• aₙ = a + (n - 1) × d

Hence,

a₂₅ = 4 + (25 - 1) × 7

a₂₅ = 4 + 24 × 7

a₂₅ = 4 + 168

a₂₅ = 172

∴ The 25th term of the A.P. is 172.

Now,

42 more than 25th term is as follows:-

= 172 + 42

= 214

∴ The number 42 more than the 25th term is 214.

Now,

We have,

• a = 4
• d = 7
• aₙ = 214

We'll now find the term of the A.P. which has 214 in it.

We already know,

aₙ = a + (n - 1) × d

Putting all the values:-

214 = 4 + (n - 1) × 7

= 214 - 4 = 7n - 7

= 210 + 7 = 7n

= 217 = 7n

=> n = 217/7

=> n = 31

∴ 31st term of the A.P. 4, 11, 18, 25, ..... is 42 more than the 25th term of this A.P.

________________________________

Answer 2

Answer:

For given , A. P.

a=4 , d= 11-4=7 and ,

$t_{25} = a + 24d$

$= > \: 4 + 24 \times 7 = 172$

$Let \: \: the \: \: required \: \: term \: \: be \: {n}^{th} \: term$

$t_{n} = t_{25} + 42 = > a + (n - 1)d = t_{25} + 42$

$= > 4 + (n - 1) \times 7 = 172 + 42$

$i.e. \: 4 + 7_{n} - 7 = 214 = > 7n = 217 \: and \: \: n = 31$

️‍️

_________________________________

THEREFORE , THE REQUIRED TERM IS

${31}^{st} \: term$

_________________________________