**Q) **Manpreet Kaur is the national record holder for women in the shot-put discipline. Her throw of 18.86m at the Asian Grand Prix in 2017 is the maximum distance for an Indian female athlete. Keeping her as a role model, Sanjitha is determined to earn gold in Olympics one day.

Initially her throw reached 7.56m only. Being an athlete in school, she regularly practiced both in the mornings and in the evenings and was able to improve the distance by 9cm every week.

During the special camp for 15 days, she started with 40 throws and every day kept increasing the number of throws by 12 to achieve this remarkable progress.

(i) How many throws Sanjitha practiced on 11th day of the camp?

(ii) What would be Sanjitha’s throw distance at the end of 6 weeks?

(or)

When will she be able to achieve a throw of 11.16 m?

(iii) How many throws did she do during the entire camp of 15 days?

**Ans: **

**VIDEO SOLUTION**

**STEP BY STEP SOLUTION**

We can clearly see that here we are given 2 APs:

AP of throw distance with a = 7.56 m and d = 9 cm = 0.09 m

AP of number of throws with a = 40 and d = 12

**(i) Throws on 11 ^{th} day:**

The throws on 11^{th} day will be given by 11^{th} term of AP of number of throws with a = 40 and d = 12

We know that the n^{th} term of an AP is given by: T_{n } = a + (n-1) d

Therefore, value of 11^{th} term, T_{11 }= 40 + (11 – 1) x 12 = 160

**Therefore, Sanjitha practiced 160 throws on 11 ^{th} day.**

**(ii) Sanjitha’s throw distance at the end of 6 weeks:**

The throw distance at end of 6 weeks will be given by 7^{th} term of AP of throw distance with a = 7.56 m and d = 0.09 m

We know that the n^{th} term of an AP is given by: T_{n } = a + (n-1) d

Therefore, value of 7^{th} term, T_{7 }= 7.56 + (7 – 1) x 0.09 = 8.10 m

**Therefore, Sanjitha’s throw distance at the end of 6 weeks will be 8.10 m**

**(or) **

**(iii) Achieving throw of 11.16 m:**

Let’s consider that the throw distance of 11.16 m will be achieved at end of n^{th} week.

We will calculate the value of n by AP of throw distance with a = 7.56 m and d = 0.09 m

We know that the n^{th} term of an AP is given by: T_{n } = a + (n-1) d

Therefore, 11.16 = 7.56 + (n – 1) x 0.09

3.60 = (n -1) x 0.09

n = 41

**Therefore, Sanjitha will ****achieve throw of 11.16 m at the end of 40 weeks.**

**(iii) Number of throws done during 15 days:**

The number of throws in 15 days will be given by sum of 15 terms of AP of number of throws with a = 40 and d = 12

We know that the sum of n terms of an AP is given by: S_{n }_{ }= (2a + (n-1) d)

Hence, sum of 15 terms S_{n }_{ }= (2 x 40 + (15 – 1) 12)

= (80 + 168) = 15 x 124 = 1860

**Therefore, Sanjitha would have done 1,860 throws during 15 days.**