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1) Algebra Formulas

2) Geometry Formulas

In the Previous Blog, we had seen some important concepts of Arithmetic & Geometric Progression. Click Here to Revise that Blog.

Anil Satpute[ Note: The following 2 files are available in the secured drive. While downloading, your email might be asked. Please provide it to download the files. I assure that your email-id will not be given to anybody ]

2) Geometry Formulas

In the Previous Blog, we had seen some important concepts of Arithmetic & Geometric Progression. Click Here to Revise that Blog.

In this Blog, we Describe the Important Formulas of an AP & GP.

1) t

_{n}= a + (n - 1) d
2) S

_{n}= n *_{ }(a + l)/2
3) S

Now we will see some examples:

Problems related to t

A) Find n

Solution:

1) Here a = 3, d = (5 - 3) = 2 so, d = 2.

2) We know that

_{n}= n *_{ }[ 2 a + (n - 1) d ) ] / 2Now we will see some examples:

Problems related to t

_{n}= a + (n - 1) d:A) Find n

^{th}^{ }term of an AP 3, 5, 7, 9 ...Solution:

1) Here a = 3, d = (5 - 3) = 2 so, d = 2.

2) We know that

t

= 3 + (n -1) (2)

= 3 + (2 n - 2)

= 1 + (2 n)

= (2 n) + 1

3) Answer: Here the n

B) Find the first term of an AP in which d = 4 and its 100

_{n}= a + (n -1) d= 3 + (n -1) (2)

= 3 + (2 n - 2)

= 1 + (2 n)

= (2 n) + 1

3) Answer: Here the n

^{th}^{ }term of an AP is t_{n}= (2 n) + 1B) Find the first term of an AP in which d = 4 and its 100

^{th}^{ }term is 403.
Solution:

1) Here d = 4 and t

2) We know that

1) Here d = 4 and t

_{100}= 403.2) We know that

t

403 = a + (100 - 1)*(4)

403 = a + (99)*(4)

403 = a + (396)

a = 403 - 396

a = 7

3) Answer: Here the 1

_{n}= a + (n -1) d403 = a + (100 - 1)*(4)

403 = a + (99)*(4)

403 = a + (396)

a = 403 - 396

a = 7

3) Answer: Here the 1

^{st}^{ }term is**a****= 7****C) If n**

^{th}^{ }term of an AP is m and m^{th}^{ }term of an AP is n, then find the value of d.
Solution:

1) Let " a " be the 1

2) We know that

1) Let " a " be the 1

^{st}^{ }term and " d " be a common difference.2) We know that

t

3) So we,

t

t

Subtract equation (2) from (1) we get,

a + (n -1) d = m

a + (m -1) d = n

(-) (-) (-)

----------------------------------

(n - 1 - m + 1) * d = (m - n)

(n - m) * d = (m - n)

d = (m - n)/(n-m)

d = - 1

4) Answer: Here common difference is

_{n}= a + (n -1) d3) So we,

t

_{n }= a + (n -1) d = m ----------- (1)t

_{m}= a + (m -1) d = n ----------- (2)Subtract equation (2) from (1) we get,

a + (n -1) d = m

a + (m -1) d = n

(-) (-) (-)

----------------------------------

(n - 1 - m + 1) * d = (m - n)

(n - m) * d = (m - n)

d = (m - n)/(n-m)

d = - 1

4) Answer: Here common difference is

**d****= - 1****Few more problems will be published in the next Blog.**