Revision: Arithmetic Progression Algebra Maths 1 SSC (English Medium) 10th Standard Maharashtra State Board

A sequence is a set of numbers arranged in a definite order.

• Each number in a sequence is called a term.

• A sequence may be written using symbols

  t₁,t₂,t₃,…,t_n
• The general term of a sequence is denoted by t_n.​

An Arithmetic Progression (A.P.) is a sequence in which the difference between consecutive terms is constant.

• Common difference = d = second term − first term

• The general form of an AP is a, a + d, a + 2d, a + 3d, …
  a = first term
  d = common difference

t_n​ = a + (n − 1)d

Used when the first term a, common difference d, and term number n are known

l = a + (n − 1)d

n^th term from end = l − (n − 1) d

Can be used only after finding:

• last term l

• number of terms

Common difference:

d = t₂ − t₁​

Nature of A.P.

• d > 0→ Increasing

• d < 0→ Decreasing

• $=0$ → Constant

If a, n, l are known → \[S_n=\frac{n}{2}(a+l)\]

If a, n, d are known → \[S_n=\frac{n}{2}\left[2a+(n-1)d\right]\]

\[1+3+\cdots+(2n-1)=n^2\]

\[1+2+\cdots+n=\frac{n(n+1)}{2}\]

\[2+4+\cdots+2n=n(n+1)\]

t_n = a + (n – 1)d

9^th term i.e., n = 9

∴ t₉ = a + (9 – 1)d

= a + 8d

It is given that t₉ = 0

∴ a + 8d = 0 ....(i)

29^th term i.e t₂₉ where n = 29

∴ t₂₉ = a + (29 – 1)d

t₂₉ = a + 28d ....(ii)

= (a + 8d) + 20d

= 0 + 20d      ......By equation (i)

∴ t₂₉ = 20d ....(ii)

t₁₉ = a + (19 – 1)d`

t₁₉ = a + 18d

= a + 8d + 10d

= 0 + 10d

t₁₉ = 10d .....(iii)

By equation (ii) and (iii)

t₂₉ = 2t₁₉

In the given problem, the 9^th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

a_n = a + (n – 1)d

We get

a₉ = a + (9 – 1)d

0 = a + 8d

a = – 8d .......(1)

Now, we need to prove that 29^th term is double of 19^th term. So, let us first find the two terms.

For 19^th term (n = 19)

a₁₉ = a + (19 – 1)d

= – 8d  + 18d    .....(Using 1)

= 10d

For 29^th term (n = 29)

a₂₉ = a + (29 – 1)d

= – 8d + 28d

= 20d

= 2 × 10d

= 2 × a₁₉  ......(Using 2)

Therefore for the given A.P. the 29^th term is double the 19^th term.

Hence proved.

Here, we are given that (m + 1)^th term is twice the (n + 1)^th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We need to prove that `a_(3m + 1) = 2a_(m + n +1)`

So, let us first find the two terms.

As we know,

`a_n = a + (n' - 1)d`

For (m + 1)^th term (n’ = m + 1)

`a_(m + 1) = a + (m + 1 - 1)d`

= a + md

For (n + 1)^th term (n’ = n + 1),

`a_(n +1) = a + (n + 1 -1)d`

= a + nd

Now, we are given that `a_(m + 1) = 2a_(n +1)`

So we get

a + md = 2(a + nd)

a + md = 2a + 2nd

md - 2nd = 2a - a

(m - 2n)d = a  ...........(1)

Further, we need to prove that the (3m + 1)^th term is twice of (m + n + 1)^th term. So let us now find these two terms,

For (m + n + 1)^th term (n' = m + n +1)

`a_(m + n + 1) = a + (m +n +1 -1)d`

= (m - 2n)d + (m + n)d

= md - 2nd + md + nd   (Using 1)

= 2md - nd

For (3m + 1)^th term (n’ = 3m + 1),

`a_(3m +1) = a + (3m + 1 -1)d`

= (m - 2n)d + 3md        (using 1)

= md - 2nd + 3md

= 4md - 2nd

= 2(2md - nd)

Therefore `a_(3m + 1) = 2a_(m + n + 1)`

Hence proved

In the given AP, let the first be a and the common difference be d.

Then, T_n = a + (n – 1)d

Now, T₄ = a + (4 – 1)d

⇒ a + 3d = 0   ...(1)

⇒ a = –3d 

Again, T₁₁ = a + (11 – 1)d

= a + 10d

= –3d + 10d

= 7d   ...[Using (1)]

Also, T₂₅ = a + (25 – 1)d

= a + 24d

= –3d + 24d

= 21d   ...[Using (1)] 

 i.e., T₂₅ = 3 × 7d = (3 × T₁₁)

Hence, 25^th term is triple its 11^th term. 

Let a be the first term and d be the common difference.

We know that, sum of first n terms = S_n = \[\frac{n}{2}\][2a + (n − 1)d]

Now,

S₄ = \[\frac{4}{2}\][2a + (4 − 1)d]

= 2(2a + 3d)

= 4a + 6d              ....(1)

S₈ = \[\frac{8}{2}\] [2a + (8 − 1)d]

= 4(2a + 7d)

= 8a + 28d            ....(2) 

S₁₂ = \[\frac{12}{2}\] [2a + (12 − 1)d]

= 6(2a + 11d)

= 12a + 66d          ....(3)

On subtracting (1) from (2), we get

S₈ − S₄ = 8a + 28d − (4a + 6d)

= 4a + 22d

Multiplying both sides by 3, we get

3(S₈ − S₄) = 3(4a + 22d)

= 12a + 66d

= S₁₂                 [From (3)]

Thus, S₁₂ = 3(S₈ − S₄).

Let a be the first term and d be the common difference.

We know that, sum of first n terms = S_n = \[\frac{n}{2}\] [2a + (n − 1)d]

Now,

S₁₀ = \[\frac{10}{2}\] [2a + (10 − 1)d]

= 5(2a + 9d)

= 10a + 45d          ....(1)

S₂₀ = \[\frac{20}{2}\] [2a + (20 − 1)d]

= 10(2a + 19d)

= 20a + 190d        ....(2) 

S₃₀ = \[\frac{30}{2}\] [2a + (30 − 1)d]

= 15(2a + 29d)

= 30a + 435d        ....(3)

On subtracting (1) from (2), we get

S₂₀ − S₁₀ = 20a + 190d − (10a + 45d)

= 10a + 145d

On multiplying both sides by 3, we get

3(S₂₀ − S₁₀) = 3(10a + 145d)

= 30a + 435d

= S₃₀                   [From (3)]

Hence, S₃₀ = 3(S₂₀ − S₁₀)